Physics > Simple Harmonic Motion > 10.0 Method for Calculating Time Period of a Simple Harmonic Motion
Simple Harmonic Motion
1.0 Types of Motion
2.0 Causes of Oscillation
3.0 Solution of the Equation of SHM
4.0 Kinematics of SHM equation i.e., $x = A\sin \left( {\omega t + \phi } \right)$
5.0 Relation between Simple Harmonic Motion and Uniform Circular Motion
6.0 Sign Convention of a Simple Harmonic Motion
7.0 How to Write the Simple Harmonic Motion Equation
8.0 Force and Energy in Simple Harmonic Motion
9.0 Basic Differential Equation of SHM
10.0 Method for Calculating Time Period of a Simple Harmonic Motion
10.1 Restoring Force or Torque Method
10.2 Energy Method
10.3 Important points
10.4 Concept of Pseudo Force
11.0 Spring Block System
12.0 Physical Pendulum
13.0 Vector Method of Combining Two or More Simple Harmonic Motions
14.0 Simple Harmonic Oscillation of a Fluid Column
10.4 Concept of Pseudo Force
10.2 Energy Method
10.3 Important points
10.4 Concept of Pseudo Force
We then use the concept of pseudo force for calculating the time period.
Case I: A simple pendulum is in a elevator moving upwards with an acceleration $\mathop a\limits^ \to $.
Using method 1:
When an elevator moves upward with an acceleration ‘$a$’, the pseudo force acts on the bob of simple pendulum is ‘$ma$’ in the downward direction.
Since the oscillatory motion is due to the tangential component of the total downward force on particle.
$${F_T} = - m\left( {g + a} \right)sin\theta $$
Note: Negative (-ve) sign indicates that the force is opposite to the displacement.
Torque about point $O$ is given by, $$\tau = - m\left( {g + a} \right)lsin\theta \quad ..\left( i \right)$$ Also, $$\tau = I\alpha = m{l^2}\alpha \quad ..\left( {ii} \right)$$ From equation $(i)$ and $(ii)$ we get, $$m{l^2}\alpha = - m\left( {g + a} \right)lsin\theta $$ For small displacement, $sin\theta \approx \theta ,$ $$\begin{equation} \begin{aligned} \alpha = \frac{{ - \left( {g + a} \right)\theta }}{l}\quad ..\left( {iii} \right) \\ \alpha = - {\omega ^2}\theta \left( {SHM\ acceleration} \right)\quad ..\left( {iv} \right) \\\end{aligned} \end{equation} $$ On equating $(iii)$ & $(iv)$ we get, $$\begin{equation} \begin{aligned} \omega = \sqrt {\frac{{g + a}}{l}} \quad ..\left( v \right) \\ T = \frac{{2\pi }}{\omega } \\\end{aligned} \end{equation} $$$$T = 2\pi \sqrt {\frac{l}{{g + a}}} \left( {{\text{from equation }}\left\{ v \right\}} \right)$$
Using Method 2:
There is another method to find the time period of a simple pendulum when placed in a elevator which is moving upward with an acceleration $a$.
To solve this type of problem find the ‘tension’ in the string at an equilibrium position.
Since the elevator is moving upward with an acceleration ‘$a$’, so the pseudo force acts in the downward direction.
From fig. SHM 15.1, we get $$\begin{equation} \begin{aligned} T = m\left( {g + a} \right) \\ \frac{T}{m} = {{\vec g}_{eff}} = \left( {g + a} \right) \\\end{aligned} \end{equation} $$
Also, time period of a simple pendulum can be written as,
$$\begin{equation} \begin{aligned} T = 2\pi \sqrt {\frac{l}{{\left| {{{\vec g}_{eff}}} \right|}}} \\ \therefore T = 2\pi \sqrt {\frac{l}{{g + a}}} \\\end{aligned} \end{equation} $$
Case II: When an elevator moves downward with an acceleration ‘$a$’, the pseudo force acts on the bob of simple pendulum is ‘$ma$’ in the upward direction direction.
Using Method 1:
When an elevator moves downward with an acceleration ‘$a$’, the pseudo force acts on the bob of simple pendulum is ‘$ma$’ in the upward direction direction.
Since the oscillatory motion is due to the tangential component of the total downward force on particle. $${F_T} = - m\left( {g - a} \right)sin\theta $$
Note: Negative (-ve) sign indicates that the force is opposite to the displacement.
Torque about point $O$ is given by,
$$\begin{equation} \begin{aligned} \tau = - m\left( {g - a} \right)lsin\theta \quad ..\left( i \right) \\ \tau = I\alpha = m{l^2}\alpha \quad ..\left( {ii} \right) \\\end{aligned} \end{equation} $$
From equation $(i)$ and $(ii)$ we get, $$m{l^2}\alpha = - m\left( {g - a} \right)lsin\theta $$
For small displacement, $sin\theta \approx \theta ,$ $$\begin{equation} \begin{aligned} \alpha = \frac{{ - \left( {g - a} \right)\theta }}{l}\quad ..\left( {iii} \right) \\ \alpha = - {\omega ^2}\theta \left( {{\text{SHM acceleration}}} \right)..\left( {iv} \right) \\\end{aligned} \end{equation} $$ On equating $(iii)$ & $(iv)$ we get,
$$\begin{equation} \begin{aligned} \omega = \sqrt {\frac{{g - a}}{l}} \quad ..\left( v \right) \\ T = 2\pi \sqrt {\frac{l}{{g - a}}} \left( {{\text{from equation }}\left\{ {\text{v}} \right\}} \right) \\\end{aligned} \end{equation} $$
Using Method 2:
There is another method to find the time period of a simple pendulum when placed in a elevator which is moving downward with an acceleration $a$.
To solve this type of problem find the ‘tension’ in the string at an equilibrium position.
Since the elevator is moving downward with an acceleration ‘$a$’, so the pseudo force acts in the upward direction.
From fig. SHM 16.1, we get
$$\begin{equation} \begin{aligned} T = m\left( {g - a} \right) \\ \frac{T}{m} = {{\vec g}_{eff}} = \left( {g - a} \right) \\\end{aligned} \end{equation} $$ Also, time period of a simple pendulum can be written as, $$\begin{equation} \begin{aligned} T = 2\pi \sqrt {\frac{l}{{\left| {{{\vec g}_{eff}}} \right|}}} \\ \therefore T = 2\pi \sqrt {\frac{l}{{g - a}}} \\\end{aligned} \end{equation} $$
Case III: When an elevator moves towards right with an acceleration ‘$a$’, the pseudo force acts on the bob of simple pendulum is ‘$ma$’ towards left direction.
So, the tension in the string is,
$$\begin{equation} \begin{aligned} T = m\sqrt {{g^2} + {a^2}} \\ \frac{T}{m} = {{\vec g}_{eff}} = \sqrt {{g^2} + {a^2}} \\\end{aligned} \end{equation} $$
Also, time period of a simple pendulum can be written as, $$\begin{equation} \begin{aligned} T = 2\pi \sqrt {\frac{l}{{\left| {{{\vec g}_{eff}}} \right|}}} \\ \therefore T = 2\pi \sqrt {\frac{l}{{\sqrt {{g^2} + {a^2}} }}} \\\end{aligned} \end{equation} $$
