what is simple pendulum experiment

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Simple Pendulum

Definition: what is a simple pendulum.

A pendulum is a device that is found in wall clocks. It consists of a weight (bob) suspended from a pivot by a string or a very light rod so that it can swing freely. When displaced to an initial angle and released, the pendulum will swing back and forth with a periodic motion. By applying Newton’s second law of motion for rotational systems, the equation of motion for the pendulum may be obtained.

Although the pendulum has a long history, Italian scientist Galileo Galilei was the first to study the properties of pendulums, beginning around 1602.

Terms Associated with Simple Pendulum

Length (L): Distance between the point of suspension to the center of the bob

Time Period (T): Time taken by the pendulum to finish one full oscillation

Linear Displacement (x): Distance traveled by the pendulum bob from the equilibrium position to one side.

Angular Displacement (θ) : The angle described by the pendulum with an imaginary axis at the equilibrium position is called the angular displacement.

Amplitude (x max ): Maximum distance traveled by the pendulum from the equilibrium position to one side before changing its direction. For angle, it is denoted by θ max .

Equation of Simple Pendulum

How to derive the formula for time period.

According to Newton’s second law,

The equation can be written in differential form as

If the amplitude of displacement is small, then the small-angle approximation holds, i.e., sin θ ~ θ.

This equation represents a simple harmonic motion. Thus, the motion of a simple pendulum is a simple harmonic motion with an angular frequency, \( \omega = \sqrt{\frac{g}{L}} \) , and linear frequency, \( f = \frac{1}{2\pi}\sqrt{\frac{g}{L}} \) . The time period is given by,

Performing dimension analysis on the right side of the above equation gives the unit of time.

[L/LT -2 ] 1/2 = [T]

The principle of a simple pendulum can be understood as follows. The restoring force of the pendulum from the above is, F = -mgL θ. This force is responsible for restoring the pendulum to its equilibrium position. However, due to the inertia of motion, the pendulum passes the equilibrium position and swings to the other side. This motion is periodic and can be solved using differential equation analysis.

After solving the differential equation, the angular displacement is given by

θ = θ max sin (ωt)

Sometimes, a phase φ is added to the above equation depending upon the initial conditions of the pendulum. Then, the equation can be written as

θ = θ max sin (ωt + φ)

A simple pendulum is a typical laboratory experiment in many academic curricula. Students are often asked to evaluate the value of the acceleration due to gravity, g, using the equation for the time period of a pendulum. Rearranging the time period equation,

Note that the component mg cos θ is balanced by the tension T of the string, i.e., T = mg cos θ.

Laws of Simple Pendulum

• Law of mass: The time period is independent of the mass of the bob.
• Law of length: The time period is directly proportional to the square root of the length.
• Law of Iscochronism: The time period is independent of the amplitude as long as the amplitude is small.
• Law of gravity: The time period is inversely proportional to the square root of the acceleration due to gravity at that place.

Uses and Applications of the Simple Pendulum

• Pendulum clock – A common household item. Every time the pendulum swings, the clock’s hand advances at a fixed rate, thus giving the time.
• Old seismometers – A pendulum with a stylus at its bottom was connected to a frame. During an earthquake, the frame moves and causes the stylus to form a pattern on paper.
• Pendulum gravimeter – A pendulum is used to measure the local gravity.
• Foucault’s pendulum – A device to measure the rotation of the earth.
• Metronome – A device used by musicians. It emits a click or a light for each beat of a predetermined interval.
• The Simple Pendulum – Acs.psu.edu
• Simple Pendulum – Hyperphysics.phy-astr.gsu.edu
• The Simple Pendulum (Simple Harmonic Motion) – Deanza.edu
• The Simple Pendulum – Iu.pressbooks.pub
• Applications of Differential Equations – Calculuslab.deltacollege.edu
• Real-world applications of Pendulums – Sites.google.com
• The Use of Pendulums in the Real World – Sciencing.com
• Oscillation of a Simple Pendulum – Acs.psu.edu
• Simple pendulum – Amrita.olabs.edu.in

Article was last reviewed on Saturday, September 30, 2023

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Simple Pendulum: Theory, Experiment, Types & Derivation

Simple Pendulum: A simple pendulum device is represented as the point mass attached to a light inextensible string and suspended from a fixed support. A simple pendulum shows periodic motion, and it occurs in the vertical plane and is mainly driven by the gravitational force.

Ever wondered why an oscillating pendulum doesn’t slow down? Or what will happen to the time period of the simple pendulum when the displacement of the bob is increased? Will it increase as the distance required to cover to complete the oscillation increases, or will it decrease as the speed at the mean position increases, or will the speed compensate for the increased distance leaving the time period unchanged? What is the difference between a physical pendulum and a simple pendulum? There are a lot of questions about the motion of a simple pendulum. Let’s read further to find out the answers.

What is Called a Simple Pendulum?

A simple pendulum is a mechanical system of mass attached to a long massless inextensible string that performs oscillatory motion. Pendulums were used to keep a track of time in ancient days. The pendulum is also used for identifying the beats.

SHM or Simple Harmonic Motion

SHM or simple harmonic motion is the type of periodic motion in which the magnitude of restoring force on the body performing SHM is directly proportional to the displacement from the mean position but the direction of force is opposite to the direction of displacement. For SHM, \(F = – K{x^n}\) The value of ‘\(n\)’ is \(1\).

Thus the acceleration of the particle is given by, \(a = \frac{F}{m}\) \(a = \frac{{ – Kx}}{m}\) Where, \(m\) is the mass of the particle. Let \({\omega ^2} = \frac{K}{m}\) As, \(\frac{K}{m}\) is a positive constant. \( \Rightarrow \,\,a = – {\omega ^2}x\) \(\omega \) is known as angular frequency of the SHM. The time period of the Simple harmonic motion is given by, \(T = \frac{{2\pi }}{\omega }\)

Following are examples of example of the simple pendulums:

Oscillating Simple Pendulum: Calculation of Time Period

It is interesting to note that the oscillation of a simple pendulum can only be considered to be a simple harmonic motion when the oscillation is small or the amplitude of oscillation is very small as compared to two lengths of the string then by using small-angle approximation the motion of a simple pendulum is considered a simple harmonic motion. When the bob is displaced by some angle then the pendulum starts the periodic motion and for small value of angle of displacement the periodic motion is simple harmonic motion with the angular displacement of the bob.

Practice Exam Questions

\(F = mg\,{\rm{sin}}\left( \theta \right)\) \(a = g\,{\rm{sin}}\left( \theta \right)\) Here \(g\) is acceleration due to gravity. For small oscillation, \(\theta \) will be small, \({\rm{sin}}\left( \theta \right) = \theta = \frac{x}{l}\) Here \(x\) is the very small linear displacement of the bob corresponding to the displaced angle. \( \Rightarrow \,\,a = g\theta \) \( \Rightarrow \,\,a = g\frac{x}{l}\) Thus the angular frequency is given by, \( \Rightarrow \,\,{\omega ^2} = \frac{g}{l}\) The time period of the pendulum is given by, \(T = \frac{{2\pi }}{\omega }\) \( \Rightarrow \,\,T = 2\pi \sqrt {\frac{l}{g}} \) Thus from the expression for a time period of a simple pendulum, we can infer that the time period does not depend on the mass of the Bob at nor varies with the change in the small amplitude of the oscillation it only depends on the length of the string and acceleration due to this property it was widely used to keep a track of fixed interval of time does it helped the musicians to be on beats

Motion of Simple Pendulum: Effect of Gravity

As the time period of simple pendulum is given by, \(T = 2\pi \sqrt {\frac{l}{g}} \) The time period of a simple pendulum is inversely proportional to the square root of acceleration due to gravity at that point. \(T \propto \frac{1}{{\sqrt g }}\) Therefore, if the acceleration due to gravity increases the time period of the simple pendulum will decrease whereas if the acceleration due to gravity decreases the time. All the simple pendulum increases.

Attempt Mock Tests

Calculation of Gravity

Acceleration due to gravity can be measured with the help of a simple experiment, The period \(T\) for a simple pendulum does not depend on the mass or the initial angular displacement but depends only on the length \(L\) of the string and the value of the acceleration due to gravity. Acceleration due to gravity is given by, \(g = \frac{{4{\pi ^2}l}}{{{T^2}}}\) One cam measure the length of the string and observe the time period and the using this formula we can find the acceleration due to gravity

Physical Pendulum

For a simple pendulum, we consider the mass of the string to be negligible as compared to the Bob but for a physical pendulum, the mass of the string need not be negligible in fact any rigid body can act as a physical pendulum.

By writing the torque equation for the rigid body about the fixed point, we get the angular acceleration of the rigid body is directly proportional to the angular displacement by using small-angle approximation. External torque on the system is zero, thus, \({\tau _{{\rm{ext}}}} = 0\)

Writing torque equation about the hinged point we get, \({\tau _0} = mgl{\rm{sin}}\left( \theta  \right) = {I_{\rm{O}}}\alpha\) Solving for \(\alpha ,\) \( \Rightarrow \,\,\,\alpha = \frac{{mgl}}{{{I_{\rm{O}}}}}{\rm{sin}}\left( \theta \right)\) Using small angle approximation, \({\rm{sin}}\left( \theta \right) = \theta \) \( \Rightarrow \,\,\,\alpha = – \frac{{mgl}}{{{I_{\rm{O}}}}}\left( \theta \right)\) Thus the angular frequency is given by, \( \Rightarrow \,\,\,{\omega ^2} = \frac{{mgl}}{{{I_{\rm{O}}}}}\) Time period of a physical pendulum is given by, \(T = 2\pi \sqrt {\frac{{{I_0}}}{{mg{l_{{\rm{cm}}}}}}} \) Where, \({I_0}\) is the moment of inertia about the fixed point trough which the axis passes. \({l_{{\rm{cm}}}}\) is the distance of the centre of mass from the axis point.

Simple Pendulum Application

Simple pendulums are used in clocks as the pendulum has a fixed time period they can be used to keep a track of time. Following are example of a simple pendulum:

Pendulums can be used as metronome.

Pendulums are used to calculate acceleration due to gravity.

Sample Problems on Real Simple Pendulum

1. A simple pendulum is suspended and the bob is subjected to a constant force in the horizontal direction. Find the time period for small oscillation.

Let the magnitude of the force be, \(F.\) Let the angle at equilibrium be, \({\theta _0}\) Let the axes be along the string and perpendicular to the string,

Balancing the forces at equilibrium, \(mg{\rm{sin}}\left( {{\theta _0}} \right) = {F_0}{\rm{cos}}\left( {{\theta _0}} \right)\) \({\rm{tan}}\left( {{\theta _0}} \right) = \frac{{{F_0}}}{{mg}}\) When the pendulum is displaced by some small angle, then,

\(F = {F_0}{\rm{cos}}\left( {\theta + {\theta _0}} \right) – mg{\rm{sin}}\left( {\theta + {\theta _0}} \right)\) \( \Rightarrow F = {F_0}\left[ {{\rm{cos}}\left( {{\theta _0}} \right){\rm{cos}}\left( \theta \right) – {\rm{sin}}\left( {{\theta _0}} \right){\rm{sin}}\left( \theta \right)} \right] – mg\left[ {{\rm{sin}}\left( {{\theta _0}} \right){\rm{cos}}\left( \theta \right) + {\rm{sin}}\left( \theta \right){\rm{cos}}\left( {{\theta _0}} \right)} \right]\) For small oscillation, \({\rm{sin}}\left( \theta \right) = \theta \) \({\rm{cos}}\left( \theta \right) = 1\) \( \Rightarrow \,\,ma = {F_0}{\rm{cos}}\left( {{\theta _0}} \right) – {F_0}{\rm{sin}}\left( {{\theta _0}} \right)\theta – mg{\rm{sin}}\left( {{\theta _0}} \right) – mg{\rm{cos}}\left( {{\theta _0}} \right)\theta \) Using, \(mg{\rm{sin}}\left( {{\theta _0}} \right) = {F_0}\rm{cos}\left( {{\theta _0}} \right)\) We get, \(a = – \frac{{\left[ {{F_0}{\rm{sin}}\left( {{\theta _0}} \right) + mg{\rm{cos}}\left( {{\theta _0}} \right)} \right]}}{m}\theta \) \( \Rightarrow \,\,\,a = – \frac{{\left[ {{F_0}{\rm{sin}}\left( {{\theta _0}} \right) + mg{\rm{cos}}\left( {{\theta _0}} \right)} \right]}}{{ml}}x\) Therefore, the angular velocity is, \({\omega ^2} = \frac{{\left[ {{F_0}{\rm{sin}}\left( {{\theta _0}} \right) + mg{\rm{cos}}\left( {{\theta _0}} \right)} \right]}}{{ml}}\) Putting in the values of \({\rm{sin}}\left( {{\theta _0}} \right)\) and \({\rm{cos}}\left( {{\theta _0}} \right)\) \(\Rightarrow \,\,\,{\omega ^2} = \frac{{\left[ {\frac{{{F_0} \times {F_0}}}{{\sqrt {{{\left( {mg} \right)}^2} + {{\left( {{F_0}} \right)}^2}} }} + \frac{{mg \times mg}}{{\sqrt {{{\left( {mg} \right)}^2} + {{\left( {{F_0}} \right)}^2}} }}} \right]}}{{ml}}\) Thus the time period will be, \(T = \frac{{2\pi }}{\omega }\) \(T = 2\pi \sqrt {\frac{l}{{\sqrt {{g^2} + {{\left( {\frac{{{F_0}}}{m}} \right)}^2}} }}} \)

2. A pendulum is hanging from the roof of a bus moving with a acceleration ‘a’. Find the time period of the pendulum.

Given, The bus is moving with the acceleration ‘\(a\)’. If we apply the concept of inertial and non-inertial frame then, a pseudo force will be applied on the bob,

Let the mass of the bob be, ‘\(m\)’. Therefore, the magnitude of the pseudo force will be, \({F_p} = ma\) The direction of the pseudo force will be in the opposite direction of the acceleration of the bus, Thus if we take the resultant acceleration experienced by the simple pendulum that is the sum of gravitation acceleration \({g_{{\rm{eff}}}} = \sqrt {{g^2} + {a^2}} \) Thus, the time period of the simple pendulum is given by, \(T = 2\pi \sqrt {\frac{l}{{{g_{{\rm{eff}}}}}}} \) Therefore, the time period is, \(T = 2\pi \sqrt {\frac{l}{{\sqrt {{g^2} + {a^2}} }}} \)

A simple pendulum is a mechanical system which consists of a light inextensible string and a small bob of some mass which is made to oscillate about its mean position from left extreme to right extreme. If the displacement of the bob is small as compared to the length of the string or the angle displaced is small then the motion can be considered to be simple harmonic motion. The total energy remains constant throughout the oscillation. The kinetic energy is maximum at the mean position whereas the potential energy is maximum at the extreme positions. The physical pendulum is a mechanical system in which a rigid body is hinged and suspended from a point. For the physical pendulum, we write the torque equation instead of force as it performs angular SHM. The Time period \(T\) for a simple pendulum does not depend on the mass or the initial angular displacement but depends only on the length \(L\) of the string and the value of the acceleration due to gravity. If the effective gravitational acceleration is changed the time period of the oscillation also changes.

FAQs on Simple Pendulum

Q What is the difference between a simple pendulum and a physical pendulum? Ans: Simple pendulum is a mechanical arrangement in which bob is suspended from a point with the help of a massless, inextensible string and performs linear simple harmonic motion for small displacement whereas a physical pendulum is a rigid body hinged from a point and is to oscillate and is performs angular simple harmonic motion for small angular displacement.

Q If a simple pendulum is moving with the acceleration ‘\(g\)’ downwards , what will be the time period of the simple pendulum hanging from its roof? Ans: The effective gravity experienced by the pendulum in this particular case will be zero thus the bob will not perform a simple harmonic motion thus the time period will not be defined as it will not have a periodic motion.

Q Is energy conserved during the oscillation of a simple pendulum? Ans: Yes, in the oscillation of a simple pendulum the total energy remains conserved while the potential and the kinetic energy keep oscillating between maxima and minima with a time period of the half to that of the oscillation of the simple pendulum.

Q What will be the time period of a simple pendulum in outer space? Ans: In outer space, there will be no gravity and thus there will be no restoring force when the pendulum will be displaced thus it will not oscillate and the will be no SHM. Thus, the tie period will not be defined.

Q What type of string should be used in a simple pendulum? Ans: The string in the simple pendulum should be inextensible that is the length of the string should not change with varying force and the mass of the string should be negligible.

We hope this detailed article on Simple Pendulum helps you in your preparation. If you get stuck do let us know in the comments section below and we will get back to you at the earliest.

Embibe wishes all the candidates the best of luck for the upcoming examination.

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Oscillatory Motion and Waves

The simple pendulum, learning objectives.

By the end of this section, you will be able to:

• Measure acceleration due to gravity.

In Figure 1 we see that a simple pendulum has a small-diameter bob and a string that has a very small mass but is strong enough not to stretch appreciably. The linear displacement from equilibrium is s , the length of the arc. Also shown are the forces on the bob, which result in a net force of − mg sin θ  toward the equilibrium position—that is, a restoring force.

Pendulums are in common usage. Some have crucial uses, such as in clocks; some are for fun, such as a child’s swing; and some are just there, such as the sinker on a fishing line. For small displacements, a pendulum is a simple harmonic oscillator. A simple pendulum is defined to have an object that has a small mass, also known as the pendulum bob, which is suspended from a light wire or string, such as shown in Figure 1. Exploring the simple pendulum a bit further, we can discover the conditions under which it performs simple harmonic motion, and we can derive an interesting expression for its period.

We begin by defining the displacement to be the arc length s . We see from Figure 1 that the net force on the bob is tangent to the arc and equals − mg sin θ . (The weight mg  has components  mg cos θ  along the string and  mg sin θ  tangent to the arc.) Tension in the string exactly cancels the component  mg  cos θ  parallel to the string. This leaves a net restoring force back toward the equilibrium position at θ  = 0.

Now, if we can show that the restoring force is directly proportional to the displacement, then we have a simple harmonic oscillator. In trying to determine if we have a simple harmonic oscillator, we should note that for small angles (less than about 15º), sin θ  ≈  θ  (sin θ  and θ differ by about 1% or less at smaller angles). Thus, for angles less than about 15º, the restoring force F is

F  ≈ − mg θ .

The displacement s is directly proportional to θ . When θ is expressed in radians, the arc length in a circle is related to its radius ( L in this instance) by  s =  L θ , so that

[latex]\theta=\frac{s}{L}\\[/latex].

For small angles, then, the expression for the restoring force is:

[latex]F\approx-\frac{mg}{L}s\\[/latex].

This expression is of the form:  F = − kx , where the force constant is given by [latex]k=\frac{mg}{L}\\[/latex] and the displacement is given by x  =  s . For angles less than about 15º, the restoring force is directly proportional to the displacement, and the simple pendulum is a simple harmonic oscillator.

Using this equation, we can find the period of a pendulum for amplitudes less than about 15º. For the simple pendulum:

[latex]\displaystyle{T}=2\pi\sqrt{\frac{m}{k}}=2\pi\sqrt{\frac{m}{\frac{mg}{L}}}\\[/latex]

Thus, [latex]T=2\pi\sqrt{\frac{L}{g}}\\[/latex] for the period of a simple pendulum. This result is interesting because of its simplicity. The only things that affect the period of a simple pendulum are its length and the acceleration due to gravity. The period is completely independent of other factors, such as mass. As with simple harmonic oscillators, the period T for a pendulum is nearly independent of amplitude, especially if θ is less than about 15º. Even simple pendulum clocks can be finely adjusted and accurate.

Note the dependence of T on g . If the length of a pendulum is precisely known, it can actually be used to measure the acceleration due to gravity. Consider Example 1.

Example 1. Measuring Acceleration due to Gravity: The Period of a Pendulum

What is the acceleration due to gravity in a region where a simple pendulum having a length 75.000 cm has a period of 1.7357 s?

We are asked to find g given the period T and the length L of a pendulum. We can solve [latex]T=2\pi\sqrt{\frac{L}{g}}\\[/latex] for g , assuming only that the angle of deflection is less than 15º.

Square [latex]T=2\pi\sqrt{\frac{L}{g}}\\[/latex] and solve for g :

[latex]g=4\pi^{2}\frac{L}{T^{2}}\\[/latex].

Substitute known values into the new equation:

[latex]g=4\pi^{2}\frac{0.750000\text{ m}}{\left(1.7357\text{ s}\right)^{2}}\\[/latex].

Calculate to find g :

g = 9.8281 m/s 2 .

This method for determining g can be very accurate. This is why length and period are given to five digits in this example. For the precision of the approximation sin θ  ≈  θ  to be better than the precision of the pendulum length and period, the maximum displacement angle should be kept below about 0.5º.

Making Career Connections

Knowing g can be important in geological exploration; for example, a map of g over large geographical regions aids the study of plate tectonics and helps in the search for oil fields and large mineral deposits.

Take Home Experiment: Determining g

Use a simple pendulum to determine the acceleration due to gravity g in your own locale. Cut a piece of a string or dental floss so that it is about 1 m long. Attach a small object of high density to the end of the string (for example, a metal nut or a car key). Starting at an angle of less than 10º, allow the pendulum to swing and measure the pendulum’s period for 10 oscillations using a stopwatch. Calculate g . How accurate is this measurement? How might it be improved?

Check Your Understanding

An engineer builds two simple pendula. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of 10 kg. Pendulum 2 has a bob with a mass of 100 kg. Describe how the motion of the pendula will differ if the bobs are both displaced by 12º.

The movement of the pendula will not differ at all because the mass of the bob has no effect on the motion of a simple pendulum. The pendula are only affected by the period (which is related to the pendulum’s length) and by the acceleration due to gravity.

PhET Explorations: Pendulum Lab

Play with one or two pendulums and discover how the period of a simple pendulum depends on the length of the string, the mass of the pendulum bob, and the amplitude of the swing. It’s easy to measure the period using the photogate timer. You can vary friction and the strength of gravity. Use the pendulum to find the value of g on planet X. Notice the anharmonic behavior at large amplitude.

Click to run the simulation.

Section Summary

• A mass m  suspended by a wire of length L  is a simple pendulum and undergoes simple harmonic motion for amplitudes less than about 15º.
• The period of a simple pendulum is [latex]T=2\pi\sqrt{\frac{L}{g}}\\[/latex], where L  is the length of the string and g  is the acceleration due to gravity.

Conceptual Questions

• Pendulum clocks are made to run at the correct rate by adjusting the pendulum’s length. Suppose you move from one city to another where the acceleration due to gravity is slightly greater, taking your pendulum clock with you, will you have to lengthen or shorten the pendulum to keep the correct time, other factors remaining constant? Explain your answer.

Problems & Exercises

As usual, the acceleration due to gravity in these problems is taken to be g  = 9.80 m/s 2 , unless otherwise specified.

• What is the length of a pendulum that has a period of 0.500 s?
• Some people think a pendulum with a period of 1.00 s can be driven with “mental energy” or psycho kinetically, because its period is the same as an average heartbeat. True or not, what is the length of such a pendulum?
• What is the period of a 1.00-m-long pendulum?
• How long does it take a child on a swing to complete one swing if her center of gravity is 4.00 m below the pivot?
• The pendulum on a cuckoo clock is 5.00 cm long. What is its frequency?
• Two parakeets sit on a swing with their combined center of mass 10.0 cm below the pivot. At what frequency do they swing?
• (a) A pendulum that has a period of 3.00000 s and that is located where the acceleration due to gravity is 9.79 m/s 2 is moved to a location where it the acceleration due to gravity is 9.82 m/s 2 . What is its new period? (b) Explain why so many digits are needed in the value for the period, based on the relation between the period and the acceleration due to gravity.
• A pendulum with a period of 2.00000 s in one location ( g = 9.80 m/s 2 ) is moved to a new location where the period is now 1.99796 s. What is the acceleration due to gravity at its new location?
• (a) What is the effect on the period of a pendulum if you double its length? (b) What is the effect on the period of a pendulum if you decrease its length by 5.00%?
• Find the ratio of the new/old periods of a pendulum if the pendulum were transported from Earth to the Moon, where the acceleration due to gravity is 1.63 m/s 2 .
• At what rate will a pendulum clock run on the Moon, where the acceleration due to gravity is 1.63 m/s 2 , if it keeps time accurately on Earth? That is, find the time (in hours) it takes the clock’s hour hand to make one revolution on the Moon.
• Suppose the length of a clock’s pendulum is changed by 1.000%, exactly at noon one day. What time will it read 24.00 hours later, assuming it the pendulum has kept perfect time before the change? Note that there are two answers, and perform the calculation to four-digit precision.
• If a pendulum-driven clock gains 5.00 s/day, what fractional change in pendulum length must be made for it to keep perfect time?

simple pendulum:  an object with a small mass suspended from a light wire or string

Selected Solutions to Problems & Exercises

7. (a) 2.99541 s; (b) Since the period is related to the square root of the acceleration of gravity, when the acceleration changes by 1% the period changes by (0.01) 2  =0.01% so it is necessary to have at least 4 digits after the decimal to see the changes.

9. (a) Period increases by a factor of 1.41 [latex]\left(\sqrt{2}\right)\\[/latex]; (b) Period decreases to 97.5% of old period

11. Slow by a factor of 2.45

13. length must increase by 0.0116%

• College Physics. Authored by : OpenStax College. Located at : http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a/College_Physics . License : CC BY: Attribution . License Terms : Located at License
• PhET Interactive Simulations . Provided by : University of Colorado Boulder . Located at : http://phet.colorado.edu . License : CC BY: Attribution

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16 Oscillatory Motion and Waves

119 16.4 The Simple Pendulum

• Measure acceleration due to gravity.

Pendulums are in common usage. Some have crucial uses, such as in clocks; some are for fun, such as a child’s swing; and some are just there, such as the sinker on a fishing line. For small displacements, a pendulum is a simple harmonic oscillator. A simple pendulum is defined to have an object that has a small mass, also known as the pendulum bob, which is suspended from a light wire or string, such as shown in Figure 1 . Exploring the simple pendulum a bit further, we can discover the conditions under which it performs simple harmonic motion, and we can derive an interesting expression for its period.

We begin by defining the displacement to be the arc length[latex]\boldsymbol{s}.[/latex]We see from Figure 1 that the net force on the bob is tangent to the arc and equals[latex]\boldsymbol{-mg\sin\theta}.[/latex](The weight[latex]\boldsymbol{mg}[/latex]has components[latex]\boldsymbol{mg\cos\theta}[/latex]along the string and[latex]\boldsymbol{mg\sin\theta}[/latex]tangent to the arc.) Tension in the string exactly cancels the component[latex]\boldsymbol{mg\cos\theta}[/latex]parallel to the string. This leaves a net restoring force back toward the equilibrium position at[latex]\boldsymbol{\theta=0}.[/latex]

Now, if we can show that the restoring force is directly proportional to the displacement, then we have a simple harmonic oscillator. In trying to determine if we have a simple harmonic oscillator, we should note that for small angles (less than about[latex]\boldsymbol{15^0}[/latex]),[latex]\boldsymbol{\sin\theta\approx\theta}[/latex]([latex]\boldsymbol{\sin\theta}[/latex]and[latex]\boldsymbol{\theta}[/latex]differ by about 1% or less at smaller angles). Thus, for angles less than about[latex]\boldsymbol{15^0},[/latex]the restoring force[latex]\boldsymbol{F}[/latex]is

The displacement[latex]\boldsymbol{s}[/latex]is directly proportional to[latex]\boldsymbol{\theta}.[/latex]When[latex]\boldsymbol{\theta}[/latex]is expressed in radians, the arc length in a circle is related to its radius ([latex]\boldsymbol{L}[/latex]in this instance) by:

For small angles, then, the expression for the restoring force is:

This expression is of the form:

where the force constant is given by[latex]\boldsymbol{k=mg/L}[/latex]and the displacement is given by[latex]\boldsymbol{x=s}.[/latex]For angles less than about[latex]\boldsymbol{15^0},[/latex]the restoring force is directly proportional to the displacement, and the simple pendulum is a simple harmonic oscillator.

Using this equation, we can find the period of a pendulum for amplitudes less than about[latex]\boldsymbol{15^0}.[/latex]For the simple pendulum:

for the period of a simple pendulum. This result is interesting because of its simplicity. The only things that affect the period of a simple pendulum are its length and the acceleration due to gravity. The period is completely independent of other factors, such as mass. As with simple harmonic oscillators, the period[latex]\boldsymbol{T}[/latex]for a pendulum is nearly independent of amplitude, especially if[latex]\boldsymbol{\theta}[/latex]is less than about[latex]\boldsymbol{15^0}.[/latex]Even simple pendulum clocks can be finely adjusted and accurate.

Note the dependence of[latex]\boldsymbol{T}[/latex]on[latex]\boldsymbol{g}.[/latex]If the length of a pendulum is precisely known, it can actually be used to measure the acceleration due to gravity. Consider the following example.

Example 1: Measuring Acceleration due to Gravity: The Period of a Pendulum

What is the acceleration due to gravity in a region where a simple pendulum having a length 75.000 cm has a period of 1.7357 s?

We are asked to find[latex]\boldsymbol{g}[/latex]given the period[latex]\boldsymbol{T}[/latex]and the length[latex]\boldsymbol{L}[/latex]of a pendulum. We can solve[latex]\boldsymbol{T=2\pi\sqrt{\frac{L}{g}}}[/latex]for[latex]\boldsymbol{g},[/latex]assuming only that the angle of deflection is less than[latex]\boldsymbol{15^0}.[/latex]

Square[latex]\boldsymbol{T=2\pi\sqrt{\frac{L}{g}}}[/latex]and solve for[latex]\boldsymbol{g}:[/latex]

[latex]\boldsymbol{g=4\pi^2}[/latex][latex]\boldsymbol{\frac{L}{T^2}}.[/latex]

Substitute known values into the new equation:

[latex]\boldsymbol{g=4\pi^2}[/latex][latex]\boldsymbol{\frac{0.75000\textbf{ m}}{(1.7357\textbf{ s})^2}}.[/latex]

Calculate to find[latex]\boldsymbol{g}:[/latex]

This method for determining[latex]\boldsymbol{g}[/latex]can be very accurate. This is why length and period are given to five digits in this example. For the precision of the approximation[latex]\boldsymbol{\sin\theta\approx\theta}[/latex]to be better than the precision of the pendulum length and period, the maximum displacement angle should be kept below about[latex]\boldsymbol{0.5^{/circ}}.[/latex]

MAKING CAREER CONNECTIONS

Knowing[latex]\boldsymbol{g}[/latex]can be important in geological exploration; for example, a map of[latex]\boldsymbol{g}[/latex]over large geographical regions aids the study of plate tectonics and helps in the search for oil fields and large mineral deposits.

TAKE-HOME EXPERIMENT: DETERMINING g

Use a simple pendulum to determine the acceleration due to gravity[latex]\boldsymbol{g}[/latex]in your own locale. Cut a piece of a string or dental floss so that it is about 1 m long. Attach a small object of high density to the end of the string (for example, a metal nut or a car key). Starting at an angle of less than[latex]\boldsymbol{10^0},[/latex]allow the pendulum to swing and measure the pendulum’s period for 10 oscillations using a stopwatch. Calculate[latex]\boldsymbol{g}.[/latex]How accurate is this measurement? How might it be improved?

Check Your Understanding

An engineer builds two simple pendula. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of[latex]\boldsymbol{10\textbf{ kg}}.[/latex]Pendulum 2 has a bob with a mass of[latex]\boldsymbol{100\textbf{ kg}}.[/latex]Describe how the motion of the pendula will differ if the bobs are both displaced by[latex]\boldsymbol{12^0}.[/latex]

PHET EXPLORATIONS: PENDULUM LAB

Play with one or two pendulums and discover how the period of a simple pendulum depends on the length of the string, the mass of the pendulum bob, and the amplitude of the swing. It’s easy to measure the period using the photogate timer. You can vary friction and the strength of gravity. Use the pendulum to find the value of[latex]\boldsymbol{g}[/latex]on planet X. Notice the anharmonic behavior at large amplitude.

Section Summary

The period of a simple pendulum is

where[latex]\boldsymbol{L}[/latex]is the length of the string and[latex]\boldsymbol{g}[/latex]is the acceleration due to gravity.

Conceptual Questions

1: Pendulum clocks are made to run at the correct rate by adjusting the pendulum’s length. Suppose you move from one city to another where the acceleration due to gravity is slightly greater, taking your pendulum clock with you, will you have to lengthen or shorten the pendulum to keep the correct time, other factors remaining constant? Explain your answer.

Problems & Exercises

As usual, the acceleration due to gravity in these problems is taken to be [latex]\boldsymbol{g=9.80\textbf{ m/s}^2},[/latex] unless otherwise specified.

1: What is the length of a pendulum that has a period of 0.500 s?

2: Some people think a pendulum with a period of 1.00 s can be driven with “mental energy” or psycho kinetically, because its period is the same as an average heartbeat. True or not, what is the length of such a pendulum?

3: What is the period of a 1.00-m-long pendulum?

4: How long does it take a child on a swing to complete one swing if her center of gravity is 4.00 m below the pivot?

5: The pendulum on a cuckoo clock is 5.00 cm long. What is its frequency?

6: Two parakeets sit on a swing with their combined center of mass 10.0 cm below the pivot. At what frequency do they swing?

7: (a) A pendulum that has a period of 3.00000 s and that is located where the acceleration due to gravity is[latex]\boldsymbol{9.79\textbf{ m/s}^2}[/latex]is moved to a location where it the acceleration due to gravity is[latex]\boldsymbol{9.82\textbf{ m/s}^2}.[/latex]What is its new period? (b) Explain why so many digits are needed in the value for the period, based on the relation between the period and the acceleration due to gravity.

8: A pendulum with a period of 2.00000 s in one location[latex]\boldsymbol{(g=9.80\textbf{ m/s}^2)}[/latex]is moved to a new location where the period is now 1.99796 s. What is the acceleration due to gravity at its new location?

9: (a) What is the effect on the period of a pendulum if you double its length?

(b) What is the effect on the period of a pendulum if you decrease its length by 5.00%?

10: Find the ratio of the new/old periods of a pendulum if the pendulum were transported from Earth to the Moon, where the acceleration due to gravity is[latex]\boldsymbol{1.63\textbf{ m/s}^2}.[/latex]

11: At what rate will a pendulum clock run on the Moon, where the acceleration due to gravity is[latex]\boldsymbol{1.63\textbf{ m/s}^2},[/latex]if it keeps time accurately on Earth? That is, find the time (in hours) it takes the clock’s hour hand to make one revolution on the Moon.

12: Suppose the length of a clock’s pendulum is changed by 1.000%, exactly at noon one day. What time will it read 24.00 hours later, assuming it the pendulum has kept perfect time before the change? Note that there are two answers, and perform the calculation to four-digit precision.

13: If a pendulum-driven clock gains 5.00 s/day, what fractional change in pendulum length must be made for it to keep perfect time?

1:  The movement of the pendula will not differ at all because the mass of the bob has no effect on the motion of a simple pendulum. The pendula are only affected by the period (which is related to the pendulum’s length) and by the acceleration due to gravity.

(a) 2.99541 s

(b) Since the period is related to the square root of the acceleration of gravity, when the acceleration changes by 1% the period changes by[latex]\boldsymbol{(0.01)^2=0.01\%}[/latex]so it is necessary to have at least 4 digits after the decimal to see the changes.

(a) Period increases by a factor of 1.41 ([latex]\boldsymbol{\sqrt{2}}[/latex])

(b) Period decreases to 97.5% of old period

Slow by a factor of 2.45

length must increase by 0.0116%.

College Physics chapters 1-17 Copyright © August 22, 2016 by OpenStax is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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Underactuated Robotics

Algorithms for Walking, Running, Swimming, Flying, and Manipulation

Russ Tedrake

Note: These are working notes used for a course being taught at MIT . They will be updated throughout the Spring 2024 semester. Lecture videos are available on YouTube .

The Simple Pendulum

Introduction.

Our goals for this chapter are modest: we'd like to understand the dynamics of a pendulum.

Why a pendulum? In part, because the dynamics of a majority of our multi-link robotics manipulators are simply the dynamics of a large number of coupled pendula. Also, the dynamics of a single pendulum are rich enough to introduce most of the concepts from nonlinear dynamics that we will use in this text, but tractable enough for us to (mostly) understand in the next few pages.

The Lagrangian derivation of the equations of motion (as described in the appendix) of the simple pendulum yields: \begin{equation*} m l^2 \ddot\theta(t) + mgl\sin{\theta(t)} = Q. \end{equation*} We'll consider the case where the generalized force, $Q$, models a damping torque (from friction) plus a control torque input, $u(t)$: $$Q = -b\dot\theta(t) + u(t).$$

Nonlinear dynamics with a constant torque

Let us first consider the dynamics of the pendulum if it is driven in a particular simple way: a torque which does not vary with time: \begin{equation} ml^2 \ddot\theta + b\dot\theta + mgl \sin\theta = u_0. \end{equation}

Simple Pendulum in Python

You can experiment with this system in using

These are relatively simple differential equations, so if I give you $\theta(0)$ and $\dot\theta(0)$, then you should be able to integrate them to obtain $\theta(t)$... right? Although it is possible, integrating even the simplest case ($b = u = 0$) involves elliptic integrals of the first kind; there is relatively little intuition to be gained here.

This is in stark contrast to the case of linear systems, where much of our understanding comes from being able to explicitly integrate the equations. For instance, for a simple linear system we have $$\dot{q} = a q \quad \rightarrow \quad q(t) = q(0) e^{at},$$ and we can immediately understand that the long-term behavior of the system is a (stable) decaying exponential if $a 0$, and that the system does nothing if $a=0$. Here we are with certainly one of the simplest nonlinear systems we can imagine, and we can't even solve this system?

All is not lost. If what we care about is the long-term behavior of the system, then there are a number of techniques we can apply. In this chapter, we will start by investigating graphical solution methods. These methods are described beautifully in a book by Steve Strogatz Strogatz94 .

The overdamped pendulum

Let's start by studying a special case -- intuitively when $b\dot\theta \gg ml^2\ddot\theta$ -- which via dimensional analysis (using the natural frequency $\sqrt{\frac{g}{l}}$ to match units) occurs when $b \sqrt\frac{l}{g} \gg ml^2$. This is the case of heavy damping, for instance if the pendulum was moving in molasses. In this case, the damping term dominates the acceleration term, and we have: $$ml^2 \ddot\theta + b\dot\theta \approx b\dot\theta = u_0 - mgl\sin\theta.$$ In other words, in the case of heavy damping, the system looks approximately first order. This is a general property of heavily-damped systems, such as fluids at very low Reynolds number.

I'd like to ignore one detail for a moment: the fact that $\theta$ wraps around on itself every $2\pi$. To be clear, let's write the system without the wrap-around as: \begin{equation}b\dot{x} = u_0 - mgl\sin{x}.\label{eq:overdamped_pend_ct}\end{equation} Our goal is to understand the long-term behavior of this system: to find $x(\infty)$ given $x(0)$. Let's start by plotting $\dot{x}$ vs $x$ for the case when $u_0=0$:

The first thing to notice is that the system has a number of fixed points or steady states , which occur whenever $\dot{x} = 0$. In this simple example, the zero-crossings are $x^* = \{..., -\pi, 0, \pi, 2\pi, ...\}$. When the system is in one of these states, it will never leave that state. If the initial conditions are at a fixed point, we know that $x(\infty)$ will be at the same fixed point.

• Locally stable in the sense of Lyapunov (i.s.L.). A fixed point, $x^*$ is locally stable i.s.L. if for every (small) $\epsilon > 0$, $\exists~\delta > 0$ such that if $\| x(0) - x^* \| < \delta$ then $\forall t$ $\| x(t) - x^*\| < \epsilon$. In words, this means that for any ball of size $\epsilon$ around the fixed point, I can create a ball of size $\delta$ which guarantees that if the system is started inside the $\delta$ ball then it will remain inside the $\epsilon$ ball for all of time.
• Locally attractive . A fixed point is locally attractive if for every (small) $\epsilon$ we have $x(0) = x^* + \epsilon$ implies that $\lim_{t\rightarrow \infty} x(t) = x^*$.
• Locally asymptotically stable . A fixed point is locally asymptotically stable if it is locally stable i.s.L. and locally attractive.
• Locally exponentially stable . A fixed point is locally exponentially stable if for every (small) $\epsilon$, we have $x(0) = x^* + \epsilon$ implies that $\| x(t) - x^* \| < Ce^{-\alpha t}$, for some positive constants $C$ and $\alpha$.
• Unstable . A fixed point is unstable if it is not locally stable i.s.L.

An initial condition near a fixed point that is stable in the sense of Lyapunov may never reach the fixed point (but it won't diverge), near an asymptotically stable fixed point will reach the fixed point as $t \rightarrow \infty$, and near an exponentially stable fixed point will reach the fixed point with a bounded rate. An exponentially stable fixed point is also an asymptotically stable fixed point, but the converse is not true. Attractivity does not actually imply Lyapunov stability † † we can't see that in one dimension so will have to hold that example for a moment , which is why we require i.s.L. specifically for the definition of asymptotic stability. Systems which are stable i.s.L. but not asymptotically stable are easy to construct (e.g. $\dot{x} = 0$). Interestingly, it is also possible to have nonlinear systems that converge (or diverge) in finite-time; a so-called finite-time stability ; we will see examples of this later in the book, but it is a difficult topic to penetrate with graphical analysis. Rigorous nonlinear system analysis is rich with subtleties and surprises. Moreover, these differences actually matter -- the code that we will write to stabilize the systems will be subtly different depending on what type of stability we want, and it can make or break the success of our methods.

Our graph of $\dot{x}$ vs. $x$ can be used to convince ourselves of i.s.L. and asymptotic stability by visually inspecting $\dot{x}$ in the vicinity of a fixed point. Even exponential stability can be inferred if we can find a negatively-sloped line passing through the equilibrium point which separates the graph of $f(x)$ from the horizontal axis. The existence of this separating line implies that the nonlinear system will converge at least as fast as the linear system represented by the straight line. I will graphically illustrate unstable fixed points with open circles and stable fixed points (i.s.L.) with filled circles.

Next, we need to consider what happens to initial conditions which begin farther from the fixed points. If we think of the dynamics of the system as a flow on the $x$-axis, then we know that anytime $\dot{x} > 0$, the flow is moving to the right, and $\dot{x} < 0$, the flow is moving to the left. If we further annotate our graph with arrows indicating the direction of the flow, then the entire (long-term) system behavior becomes clear:

For instance, we can see that any initial condition $x(0) \in (-\pi,\pi)$ will result in $\lim_{t\rightarrow \infty} x(t) = 0$. This region is called the region of attraction (or basin of attraction ) of the fixed point at $x^* = 0$. Basins of attraction of two fixed points cannot overlap, and the manifold separating two basins of attraction is called the separatrix . Here the unstable fixed points, at $x^* = \{.., -\pi, \pi, 3\pi, ...\}$ form the separatrix between the basins of attraction of the stable fixed points. In general, regions of attraction are always open, connected, invariant sets, with boundaries formed by trajectories.

As these plots demonstrate, the behavior of a first-order one dimensional system on a line is relatively constrained. The system will either monotonically approach a fixed-point or monotonically move toward $\pm \infty$. There are no other possibilities. Oscillations, for example, are impossible. Graphical analysis is a fantastic analysis tool for many first-order nonlinear systems (not just pendula); as illustrated by the following example:

Nonlinear autapse

These equations are not arbitrary - they are actually a model for one of the simplest neural networks, and one of the simplest models of persistent memory Seung00 . In the equation $x$ models the firing rate of a single neuron, which has a feedback connection to itself. $\tanh$ is the activation (sigmoidal) function of the neuron, and $w$ is the weight of the synaptic feedback.

Experiment with it for yourself:

As I've also provided the equations of an LSTM unit that you can also experiment with. See if you can figure it out!

Architectures like the Transformer for sequence modeling can also be understood through the lens of dynamical systems theory (as autoregressive models), but we will defer that discussion until later in the notes.

One last piece of terminology. In the neuron example, and in many dynamical systems, the dynamics were parameterized; in this case by a single parameter, $w$. As we varied $w$, the fixed points of the system moved around. In fact, if we increase $w$ through $w=1$, something dramatic happens - the system goes from having one fixed point to having three fixed points. This is called a bifurcation . This particular bifurcation is called a pitchfork bifurcation. We often draw bifurcation diagrams which plot the fixed points of the system as a function of the parameters, with solid lines indicating stable fixed points and dashed lines indicating unstable fixed points, as seen in the figure:

Our pendulum equations also have a (saddle-node) bifurcation when we change the constant torque input, $u_0$. Finally, let's return to the original equations in $\theta$, instead of in $x$. Only one point to make: because of the wrap-around, this system will appear to have oscillations. In fact, the graphical analysis reveals that the pendulum will turn forever whenever $|u_0| > mgl$, but now you understand that this is not an oscillation, but an instability with $\theta \rightarrow \pm \infty$.

You can find another beautiful example of these concepts (fixed points, basins of attraction, bifurcations) from recurrent neural networks in the exercise below on Hopfield networks .

The undamped pendulum with zero torque

Consider again the system $$ml^2 \ddot\theta = u_0 - mgl \sin\theta - b\dot\theta,$$ this time with $b = 0$. This time the system dynamics are truly second-order. We can always think of any second-order system, $$\ddot{\theta} = f(\theta,\dot\theta),$$ as (coupled) first-order system with twice as many variables. This system is equivalent to the two-dimensional first-order system \begin{align*} \dot x_1 =& x_2 \\ \dot x_2 =& f(x_1,x_2), \end{align*} where $x_1 = \theta$ and $x_2 = \dot \theta$. Therefore, the graphical depiction of this system is not a line, but a vector field where the vectors $[\dot x_1, \dot x_2]^T$ are plotted over the domain $(x_1,x_2)$. This vector field is known as the phase portrait of the system.

In this section we restrict ourselves to the simplest case when $u_0 = 0$. Let's sketch the phase portrait. First sketch along the $\theta$-axis. The $x$-component of the vector field here is zero, the $y$-component is $-\frac{g}{l}\sin\theta.$ As expected, we have fixed points at $\pm \pi, ...$ Now sketch the rest of the vector field. Can you tell me which fixed points are stable? Some of them are stable i.s.L., none are asymptotically stable.

Orbit calculations

You might wonder how we drew the black contour lines in the figure above. We could have obtained them by simulating the system numerically, but those lines can be easily obtained in closed-form. Directly integrating the equations of motion is difficult, but at least for the case when $u_0 = 0$, we have some additional physical insight for this problem that we can take advantage of. The kinetic energy, $T$, and potential energy, $U$, of the pendulum are given by $$T = \frac{1}{2}I\dot\theta^2, \quad U = -mgl\cos(\theta),$$ where $I=ml^2$ and the total energy is $E(\theta,\dot\theta) = T(\dot\theta)+U(\theta)$. The undamped pendulum is a conservative system: total energy is a constant over system trajectories. Using conservation of energy, we have: \begin{gather*} E(\theta(t),\dot\theta(t)) = E(\theta(0),\dot\theta(0)) = E_0 \\ \frac{1}{2} I \dot\theta^2(t) - mgl\cos(\theta(t)) = E_0 \\ \dot\theta(t) = \pm \sqrt{\frac{2}{I}\left[E_0 + mgl\cos\left(\theta(t)\right)\right]} \end{gather*} Using this, if you tell me $\theta$ I can determine one of two possible values for $\dot\theta$, and the solution has all of the richness of the black contour lines from the plot. This equation has a real solution when $\cos(\theta) > \cos(\theta_{max})$, where $$\theta_{max} = \begin{cases} \cos^{-1}\left( -\frac{E_0}{mgl} \right), & E_0 < mgl \\ \pi, & \text{otherwise}. \end{cases}$$ Of course this is just the intuitive notion that the pendulum will not swing above the height where the total energy equals the potential energy. As an exercise, you can verify that differentiating this equation with respect to time indeed results in the equations of motion.

The particular orbit defined by $E = mgl$ is special -- this is the orbit that visits the (unstable) equilibrium at the upright. This is the homoclinic orbit of the pendulum.

Trajectory calculations

For completeness, I'll include what it would take to solve for $\theta(t)$, even thought it cannot be accomplished using elementary functions. Feel free to skip this subsection. We begin the integration with \begin{gather*} \frac{d\theta}{dt} = \sqrt{\frac{2}{I}\left[E + mgl\cos\left(\theta(t)\right)\right]} \\ \int_{\theta(0)}^{\theta(t)} \frac{d\theta}{\sqrt{\frac{2}{I}\left[E + mgl\cos\left(\theta(t)\right)\right]}} = \int_0^t dt' = t \end{gather*} The integral on the left side of this equation is an (incomplete) elliptic integral of the first kind. Using the identity: $$\cos(\theta) = 1 - 2 \sin^2(\frac{1}{2}\theta),$$ and manipulating, we have $$t = \sqrt{\frac{I}{2(E+mgl)}} \int_{\theta(0)}^{\theta(t)} \frac{d\theta}{\sqrt{1 - k_1^2\sin^2(\frac{\theta}{2})}}, \quad \text{with }k_1=\sqrt{\frac{2mgl}{E+mgl}}.$$ In terms of the incomplete elliptic integral function, $$F(\phi,k) = \int_0^\phi \frac{d\theta}{\sqrt{1-k^2\sin^2\theta}},$$ accomplished by a change of variables. If $E