Question

(i) Amplitude of particle performing SHM is $20\,cm$ its maximum velocity during oscillation is $150\,cm\,{\sec ^{ - 1}}$ find its displacement when its velocity is $90\,cm\,{\sec ^{ - 1}}$
(ii) A particle performing SHM has maximum velocity $9.6\,cm\,{\sec ^{ - 1}}$ path length is $12\,cm$ find its period.

Answer 1

In order to solve this question we need to understand Simple Harmonic Motion. Simple harmonic motion is periodic motion which oscillates around mean position and it repeats itself after some time. In this motion the force causing displacement of the body is directly proportional to displacement but in the opposite direction of motion. In this motion energy is always conserved, so there is oscillation of kinetic and potential energy around the mean point but their sum is always constant in nature. Kinetic energy is maximum at mean position while the potential energy is maximum at extreme positions.

Complete step by step answer:
Consider a particle performing Simple harmonic Motion.Let its displacement be given by x and let it oscillate with angular velocity $\omega$. So displacement is given by,
$x = A\sin (\omega t) \to (i)$
Here, $A$ is amplitude or the max displacement of the body performing SHM.
So velocity is given by, $v = \dfrac{{dx}}{{dt}}$ and $\dfrac{d}{{dt}}(\sin \omega t) = \omega \cos \omega t$
So, $v = A\omega \cos (\omega t) \to (iii)$
Since $\cos (\omega t) = \sqrt {(1 - {{\sin }^2}(\omega t))} \to (ii)$
And from (i) we get, $\sin (\omega t) = \dfrac{x}{A}$
Putting value in (ii) we get,
$\cos (\omega t) = \sqrt {1 - \dfrac{{{x^2}}}{{{A^2}}}}$
$\Rightarrow \cos (\omega t) = \dfrac{1}{A}\sqrt {{A^2} - {x^2}}$
Putting value in (iii) we get,
$v = A\omega (\dfrac{1}{A}\sqrt {{A^2} - {x^2}} )$
$\Rightarrow v = \omega \sqrt {{A^2} - {x^2}}$
So maximum velocity is at $x = 0$ so max velocity is, ${v_{\max }} = A\omega$

(i) According to given question, $A = 20cm$ and ${v_{\max }} = 150\,cm\,{\sec ^{ - 1}}$
Putting values we get, $\omega = \dfrac{{{v_{\max }}}}{A}$
$\omega = \dfrac{{150}}{{20}}$
$\Rightarrow \omega = 7.5\,rad\,{\sec ^{ - 1}}$
So displacement at velocity $90cm{\sec ^{ - 1}}$ is given by,
$v = \omega \sqrt {{A^2} - {x^2}}$
$\Rightarrow {v^2} = {\omega ^2}({A^2} - {x^2})$
$\Rightarrow {x^2} = {A^2} - \dfrac{{{v^2}}}{{{\omega ^2}}}$
$\Rightarrow x = \sqrt {{A^2} - \dfrac{{{v^2}}}{{{\omega ^2}}}}$
Putting values we get, $x = \sqrt {({{20}^2}) - \dfrac{{({{90}^2})}}{{({{7.5}^2})}}}$
$x = \sqrt {400 - 144}$
$\Rightarrow x = \sqrt {256}$
$\therefore x = 16\,cm$

So displacement at velocity $90\,cm\,{\sec ^{ - 1}}$ is $16\,cm$.

(ii) Path length is, $2A = 12\,cm$.
$A = 6\,cm$ and maximum velocity is, ${v_{\max }}9.6cm{\sec ^{ - 1}}$.
Also maximum velocity is, ${v_{\max }} = \omega A$.
So, angular velocity is $\omega = \dfrac{{{v_{\max }}}}{A}$
Putting values we get, $\omega = \dfrac{{9.6}}{6}$
$\omega = 1.6\,rad\,{\sec ^{ - 1}}$
Since time period is given by,
$T = \dfrac{{2\pi }}{\omega }$
Putting values we get,
$T = \dfrac{{2\pi }}{{1.6}}$
$\therefore T = 3.927\sec$

So the period is $3.927\sec$.

Note: It should be remembered that all harmonic motions are periodic motions but all periodic motions are not harmonic motion. As in harmonic motion applied force is directly proportional to displacement but it acts in opposite direction of displacement but this is not a necessity in all periodic motions. Examples of harmonic motion are the motion of pendulum, motion of spring etc.