JEE Main & Advanced Physics Simple Harmonic Motion How to Find Frequency and Time Period of S.H.M.

Step 1 : When particle is in its equilibrium position, balance all forces acting on it and locate the equilibrium position mathematically.

Step 2 : From the equilibrium position, displace the particle slightly by a displacement y and find the expression of net restoring force on it.

Step 3 : Try to express the net restoring force acting on particle as a proportional function of its displacement from mean position. The final expression should be obtained in the form.

\[F=-\,ky\]

Here we put \[-ve\] sign as direction of F is opposite to the displacement y. If a be the acceleration of particle at this displacement, we have \[a=-\,\left( \frac{k}{m} \right)\,y\]    

Step 4 : Comparing this equation with the basic differential equation of S.H.M. we get \[{{\omega }^{2}}=\frac{k}{m}\]\[\Rightarrow \]\[\omega =\sqrt{\frac{k}{m}}\] or \[n=\frac{1}{2\pi }\sqrt{\frac{k}{m}}\]

As \[\omega \] is the angular frequency of the particle in S.H.M., its time period of oscillation can be given as \[T=\frac{2\pi }{\omega }\]\[=2\pi \sqrt{\frac{m}{k}}\]

(i) In different types of S.H.M. the quantities m and k will go on taking different forms and names. In general m is called inertia factor and k is called spring factor.

Thus \[T=2\pi \sqrt{\frac{\text{Inertia factor}}{\text{Spring factor }}}\]or \[n=\frac{1}{2\pi }\sqrt{\frac{\text{Spring}\,\text{factor}}{\text{Inertia factor}}}\]

(ii) In linear S.H.M. the spring factor stands for force per unit displacement and inertia factor for mass of the body executing S.H.M. and in Angular S.H.M. k stands for restoring torque per unit angular displacement and inertial factor for moment of inertia of the body executing S.H.M.

For linear S.H.M.

\[T=2\pi \sqrt{\frac{m}{k}}=\sqrt{\frac{m}{\text{Force/Displacement}}}\]\[=2\pi \sqrt{\frac{\,\text{Displacement}}{\text{Acceleration}}}\]