pendulum simple harmonic motion experiment

Required Practical: Investigating SHM ( AQA A Level Physics )

Revision note.

Physics Project Lead

Required Practical: Investigating SHM

Equipment list.

• Stopwatch = ±0.01 s
• Metre Ruler = ±1 mm

SHM in a Mass-Spring System

• This experiment aims to calculate the spring constant of a spring in a mass-spring system
• This is just one example of how this required practical might be carried out
• Independent variable = mass,  m
• Dependent variable = time period,  T
• Spring constant,  k
• Number of oscillations

The setup of apparatus to detect oscillations of a mass-spring system

• Set up the apparatus, with no masses hanging on the holder to begin with (just the 100 g mass attached to it)
• Pull the mass hanger vertically downwards between 2-5 cm as measured from the ruler and let go. The mass hanger will begin to oscillate
• Start the stopwatch when it passes the nail marker
• Stop the stopwatch after 10 complete oscillations and record this time. Divide the time by 10 for the time period (which is the mean)
• Add a 50 g mass to the holder and repeat the above between 8-10 readings. Make sure the mass is pulled down by the same length before letting go
• An example table might look like this:

 Analysing the Results

• Obtain an equation for the spring constant,  k 
• Then plot a suitable graph to obtain a value for k
• Start with the time period of a mass-spring system from the equation:

• T  = time period (s)
• m = mass (kg)
• k = spring constant (N m –1 )
• Squaring both sides of the equation gives:

• Gradient = 4π 2 / k
• The spring constant, k , is therefore equal to:

• Where  T 2 and  m  are directly proportional to each other
• The graph is a straight line with a positive gradient

• Where k is found from the gradient of a force F  extension x  graph

SHM in a Simple Pendulum

• This experiment aims to calculate the acceleration due to gravity of a simple pendulum
• Independent variable = length, L
• Mass of pendulum bob,  m

• Set up the apparatus, with the length of the pendulum at 0.2 m
• Make sure the pendulum hangs vertically downwards at equilibrium and inline directly in front of the needle marker
• Pull the pendulum to the side at a very small angle then let go. The pendulum will begin to oscillate
• Start the stopwatch when the pendulum passes the needle marker in its equilibrium. One complete oscillation occurs when the pendulum passes through the equilibrium, to one maximum and then the other, and back to the equilibrium again (not just from side to side)
• Stop the stopwatch after 10 complete oscillations and record the total time. Divide the time by 10 to obtain the time period (which is the mean)
• Adjust the string to increase the length of the pendulum and the wooden block. Repeat the above for 8-10 readings. The ruler is used to measure the string. Ensure it is measured from the wooden blocks to the centre of mass of the bob.
• Oscillations should be counted as follows:

Analysing the Results

• Obtain an equation for the   acceleration due to gravity, g 
• Then plot a suitable   graph   to obtain a value for g
• The time period of a simple pendulum is given by:

• T = time period (s)
• L = length of the pendulum (m)
• g = acceleration due to gravity (m s –2 )
• Squaring both sides of the equation gives

• gradient m = 4π 2 / g

The acceleration due to gravity is equal to:

• Where  T 2   and  L are   directly proportional   to each other
• The graph is a   straight line   with a   positive gradient

• The accuracy of the experiment can be determined by comparing the obtained value of  g  to the accepted value of acceleration due to gravity,  g = 9.81 m s −2

Evaluating the Experiments

Systematic Errors :

• Reduce parallax error by viewing the marker at eye level

Random Errors :

• Record the time taken for 10 full oscillations, then divide by 10 for one period, to reduce random errors
• For the simple pendulum, the oscillations may not completely go from side to side, and the object may move in a circle. Therefore, keep the amplitudes of oscillation relatively small (only a few cm) and repeat any readings that take a different trajectory
• The equation for the time period of a pendulum bob only works for small angles , so make sure the pendulum is not pulled too far out to the side for the oscillation
• For the mass-spring system, the oscillations may not stay completely vertical. Therefore, keep the amplitudes relatively small (only a few cm) and repeat the readings making sure they are vertical
• When setting an oscillation in motion make sure the mass is pulled to the side by the same angle every time 
• A motion tracker and data logger could provide a more accurate value for the time period and reduce the random errors in starting and stopping the stopwatch (due to reflex times)

Safety Considerations

• Place a soft surface directly below the equipment to reduce the damage caused by a falling pendulum or spring
• Only pull down the mass and spring system a few centimetres for the oscillations, as larger oscillations could cause the masses to fall off and damage the equipment
• The wooden blocks must be tightly clamped together to hold the string for the pendulum in place, otherwise, the pendulum may dislodge during oscillations and fall off

Worked example

A student investigates the relationship between the time period and the mass of a mass-spring system that oscillates with simple harmonic motion. They obtain the following results:

Calculate the value of the spring constant of the spring used in this experiment.

Step 1: Complete the table

Add the extra column T 2 and calculate the values

Step 2: Plot the graph of T 2 against the mass m

Make sure the axes are properly labelled and the line of best fit is drawn with a ruler.

The line of best fit should have an equal number of points above and below it.

Step 3: Calculate the gradient of the graph

The gradient is calculated by:

Step 4: Calculate the spring constant, k

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16.4 The Simple Pendulum

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 16.14 . 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 s size 12{s} {} . We see from Figure 16.14 that the net force on the bob is tangent to the arc and equals − mg sin θ − mg sin θ size 12{ - ital "mg""sin"θ} {} . (The weight mg mg size 12{ ital "mg"} {} has components mg cos θ mg cos θ size 12{ ital "mg""cos"θ} {} along the string and mg sin θ mg sin θ size 12{ ital "mg""sin"θ} {} tangent to the arc.) Tension in the string exactly cancels the component mg cos θ mg cos θ size 12{ ital "mg""cos"θ} {} parallel to the string. This leaves a net restoring force back toward the equilibrium position at θ = 0 θ = 0 size 12{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 15º size 12{"15"°} {} ), sin θ ≈ θ sin θ ≈ θ size 12{"sin"θ approx θ} {} ( sin θ sin θ size 12{"sin"θ} {} and θ θ size 12{θ} {} differ by about 1% or less at smaller angles). Thus, for angles less than about 15º 15º size 12{"15"°} {} , the restoring force F F size 12{F} {} is

The displacement s s size 12{s} {} is directly proportional to θ θ size 12{θ} {} . When θ θ size 12{θ} {} is expressed in radians, the arc length in a circle is related to its radius ( L L size 12{L} {} 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 k = mg / L k = mg / L and the displacement is given by x = s x = s size 12{x=s} {} . For angles less than about 15º 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 15º . 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 T T size 12{T} {} for a pendulum is nearly independent of amplitude, especially if θ θ size 12{θ} {} is less than about 15º 15º size 12{"15"°} {} . Even simple pendulum clocks can be finely adjusted and accurate.

Note the dependence of T T size 12{T} {} on g g size 12{g} {} . 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 16.5

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 g size 12{g} {} given the period T T size 12{T} {} and the length L L size 12{L} {} of a pendulum. We can solve T = 2π L g T = 2π L g size 12{T=2π sqrt { { {L} over {g} } } } {} for g g size 12{g} {} , assuming only that the angle of deflection is less than 15º 15º size 12{"15º"} {} .

• Square T = 2π L g T = 2π L g size 12{T=2π sqrt { { {L} over {g} } } } {} and solve for g g size 12{g} {} : g = 4π 2 L T 2 . g = 4π 2 L T 2 . size 12{g=4π rSup { size 8{2} } { {L} over {T rSup { size 8{2} } } } } {} 16.30
• Substitute known values into the new equation: g = 4π 2 0 . 75000 m 1 . 7357 s 2 . g = 4π 2 0 . 75000 m 1 . 7357 s 2 . size 12{g=4π rSup { size 8{2} } { {0 "." "75000"" m"} over { left (1 "." "7357 s" right ) rSup { size 8{2} } } } } {} 16.31
• Calculate to find g g size 12{g} {} : g = 9 . 8281 m / s 2 . g = 9 . 8281 m / s 2 . size 12{g=9 "." "828"m/s rSup { size 8{2} } } {} 16.32

This method for determining g g 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 sin θ ≈ θ size 12{"sin"θ approx θ} {} to be better than the precision of the pendulum length and period, the maximum displacement angle should be kept below about 0.5º 0.5º size 12{0 "." 5°} {} .

Making Career Connections

Knowing g g g can be important in geological exploration; for example, a map of g g size 12{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 g g

Use a simple pendulum to determine the acceleration due to gravity g g 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 10º size 12{"10"°} {} , allow the pendulum to swing and measure the pendulum’s period for 10 oscillations using a stopwatch. Calculate g g size 12{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 10 kg size 12{"10"`"kg"} {} . Pendulum 2 has a bob with a mass of 100 kg 100 kg size 12{"100"`"kg"} {} . Describe how the motion of the pendula will differ if the bobs are both displaced by 12º 12º size 12{"12"°} {} .

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 g on planet X. Notice the anharmonic behavior at large amplitude.

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Pendulum Lab

• Periodic Motion
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Description

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, the strength of gravity, and the amplitude of the swing. Observe the energy in the system in real-time, and vary the amount of friction. Measure the period using the stopwatch or period timer. Use the pendulum to find the value of g on Planet X. Notice the anharmonic behavior at large amplitude.

Sample Learning Goals

• Design experiments to determine which variables affect the period of a pendulum
• Quantitatively describe how the period of a pendulum depends on these variables
• Explain the small-angle approximation, and define what constitutes a “small” angle
• Determine the gravitational acceleration of Planet X
• Explain the conservation of mechanical energy, using kinetic energy and gravitational potential energy
• Describe the Energy Graph from the position and speed of the pendulum
• Acceleration
• Simple Harmonic Motion
• Harmonic Motion
• Kinetic Energy
• Potential Energy
• Thermal Energy
• Air Resistance

For Teachers

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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%

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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.

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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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