Simple Harmonic Motion
4.0 Kinematics of SHM equation i.e., $x = A\sin \left( {\omega t + \phi } \right)$
4.0 Kinematics of SHM equation i.e., $x = A\sin \left( {\omega t + \phi } \right)$
- $\sin \left( {\omega t + \phi } \right)$ varies from $+1$ to $–1$. So, the SHM equation varies from $+A$ to $–A$. Thus the range of vibration is $2A$.
- $\omega$, is known as the ‘angular frequency’. $\phi$ is known as the ‘phase constant’ or ‘phase angle’, whose value depends on the initial conditions.
- When the body starts oscillating from equilibrium position, i.e. $x = 0$ at $t = 0$, then $\phi = 0$.
- Time period ($T$) of SHM or time taken to complete one vibration is independent of the phase constant ($\phi$). $$T = \frac{{2\pi }}{\omega } = 2\pi \sqrt {\frac{m}{k}} $$ where $T$ is called the time period of oscillation.
- Frequency ($f$), the number of vibrations per second, $$f = \frac{1}{T} = \frac{1}{{2\pi }}\sqrt {\frac{k}{m}} $$
- The velocity of particle is given by, $$v = \frac{{dx}}{{dt}} = \omega A\cos \left( {\omega t + \phi } \right)$$ he velocity of the particle varies periodically between $ + \omega A$ and $ - \omega A$. $$v = \frac{{dx}}{{dt}} = \omega \sqrt {{A^2} - {x^2}} ,[As\sqrt {{A^2} - {x^2}} = A\cos \left( {\omega t + \phi } \right)]$$
Conclusion: A particle in SHM has maximum velocity at mean position and zero velocity at the extreme position.
- The acceleration of the particle is given by, $$a = \frac{{dv}}{{dt}} = - {\omega ^2}A\sin \left( {\omega t + \phi } \right) = - {\omega ^2}x$$ The acceleration of the particle varies periodically between $ + {\omega ^2}A$ and $ - {\omega ^2}A$.
Conclusion: A particle in simple harmonic motion has zero acceleration at the mean position and maximum acceleration at the extreme position.
