Question

For a shm the time period is 2s. The displacement from the mean position is -10 m. Calculate the instantaneous acceleration?

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Accleration=10\pi^{2} rad/{s}^{2}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Time \: period(T) = 2 \: s \\ \\ \tt: \implies Displacement(x)= - 10 \: m \\ \\\red{\underline \bold{To \: Find :}} \\ \tt: \implies Instantaneous \: acceleration = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies T= \frac{2\pi}{ \omega} \\ \\ \tt: \implies 2 = \frac{2\pi}{ \omega} \\ \\ \tt: \implies \omega = \frac{2\pi}{2} \\ \\ \green{\tt: \implies \omega = \pi \: rad/s} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies Acceleration = - \omega^{2} x \\ \\ \tt: \implies Acceleration = - \pi^{2} ( - 10) \\ \\ \green{\tt: \implies Acceleration = 10 \pi^{2} rad/{s}^{2} }$

Answer 2

GiveN :-

$\textsf{Time period for a SHM(T) = 2 s}$

$\textsf{Displacement from the mean position(x) = -10 m}$

TO FinD :-

$\textsf{The instantaneous acceleration(a) of the SHM.}$

AcknowledgemenT :-

$\bullet\ \rm{T=\dfrac{2\pi}{\omega}}$

$\bullet\ \rm{Acceleration(a)=- \omega^2x}$

SolutioN :-

$\textsl{Putting the values :-}$

$\sf{T = \dfrac{2\pi}{\omega}}\\\\\sf{\longrightarrow 2= \dfrac{2\pi}{\omega}}\\\\\sf{\longrightarrow \omega=\dfrac{2\pi}{2}}\\\\\boxed{\sf{\longrightarrow \omega=\pi\ rad/sec}}$

$\rule{130}{2}$

$\textsl{Now,}$

$\sf{a=-\omega^2x}\\\\\sf{\longrightarrow a=-(\pi)^2\times-10}\\\\ \sf{\longrightarrow a=-\pi^2 (-10)}\\\\ \underline{\boxed{\sf{\longrightarrow a=10\pi^2\ rad/sec^2}}}$                  

Therefore, the instantaneous acceleration is 10π² rad/sec².