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
10.1 Restoring Force or Torque Method
10.2 Energy Method
10.3 Important points
10.4 Concept of Pseudo Force
Step 1.
Find the stable equilibrium position, which is usually known, as the mean position. Net force or torque on the particle in this position is zero. Potential energy is minimum.
Step 2.
Displace the particle from its mean position by a small displacement $x$ (in case of linear SHM) or $\theta$ (in case of an angular SHM).
Step 3.
Find net force or torque in the displaced position.
Step 4.
Show that this force or torque has tendency to bring the particle back to its mean position and magnitude of force or torque is proportional to displacement, i.e., $$F \propto - x\quad or\quad F = - kx\quad ..\left( i \right)$$ or, $$\tau \propto - \theta \quad or\quad \tau = - k\theta \quad ..\left( {ii} \right)$$ This force or torque is also known as restoring force or restoring torque.
Step 5.
Find linear acceleration by dividing Eq. (i) by mass $m$ or angular acceleration by dividing Eq. (ii) by moment of inertia $I$. Hence, $$a = - \frac{k}{m}x = - {\omega ^2}x\quad or\quad \alpha = - \frac{k}{I}\theta = - {\omega ^2}\theta $$
Step 6.
Finally,
$$\begin{equation} \begin{aligned} \omega = \sqrt {\frac{k}{m}} \quad or\quad \sqrt {\frac{k}{I}} \\ \frac{{2\pi }}{T} = \sqrt {\frac{k}{m}} \quad or\quad \sqrt {\frac{k}{I}} \\ T = 2\pi \sqrt {\frac{k}{I}} \quad or\quad 2\pi \sqrt {\frac{k}{I}} \\\end{aligned} \end{equation} $$
Example: The Simple Pendulum (Using Restoring Force or Torque Method)
An example of SHM is the motion of a pendulum. A simple pendulum is defined as a particle of mass $m$ suspended from a point $O$ by a string of length $l$ and of negligible mass.
When the particle is pulled aside to position $S$, so that the string makes and angle ${\theta _o}$ with the vertical $OR$ and then released, the pendulum will oscillate between $S$ and symmetric position $Q$. The oscillatory motion is due to the tangential component $F_T$ of the weight $mg$ of the particle. The force $F_T$ is maximum at $S$ and $Q$, and zero at $R$. Thus, we can write, $${F_T} = - mgsin\theta $$ Here minus sign appears because it is opposite to the displacement.
$$\begin{equation} \begin{aligned} x = CA \\ m{a_T} = - mgsin\theta \\ {a_T} = l\alpha \left( {{\text{Relation between linear and angular acceleration}}} \right) \\ sin\theta \approx \theta \left( {{\text{for small oscillations}}} \right) \\ ml\alpha = - mg\theta \\ \alpha = - \left( {\frac{g}{l}} \right)\theta \quad ..\left( A \right) \\ \alpha = - {\omega ^2}\theta \left( {{\text{SHM acceleration}}} \right)\quad ..\left( B \right) \\\end{aligned} \end{equation} $$ On equating $A$ & $B$ we get, $$\omega = \sqrt {\frac{g}{l}} \quad ..\left( C \right)$$$$T = \frac{{2\pi }}{\omega }$$$$T = 2\pi \sqrt {\frac{l}{g}} {\text{ }}\left( {{\text{from equation C}}} \right)$$
Note that the period is independent of the mass of the pendulum.
