Physics, asked by nickyraptor6262, 9 days ago

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

Answers

Answered by BrainlyConqueror0901
33

\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} }

Answered by CunningKing
20

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.

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