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.

It is interesting to note that the time period of the real simple pendulum remains constant even if the amplitude is changed but if the acceleration due to gravity changes the time period of the simple pendulum also changes. This property is also used to determine or verify acceleration due to gravity. The time period also depends on the length of the string to which the Bob of the simple pendulum is attached.

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:

• Clock
• Metronome
• Bob

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,
\(T = 2\pi \sqrt {\frac{l}{{\sqrt {{g^2} + {a^2}} }}} \)

Summary

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.

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