A particle executes simple harmonic motion with a time period $0.6 \mathrm{~s}$ and amplitude $10 \mathrm{~cm}$. Then the mean velocity of the particle over the time interval during which it travels a distance $5 \mathrm{~cm}$ starting from the equilibrium position.

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Aicharya · Accepted Answer

We have $\begin{aligned} & x=A \sin \omega t \Rightarrow 5=10 \sin \omega t \ & \Rightarrow \frac{1}{2}=\sin \omega t \Rightarrow t=\frac{\pi}{6 \omega}=\frac{\pi 	imes T}{6 	imes 2 \pi}=\frac{T}{12} \end{aligned}$ Then, $V=\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{A} \omega \cos \omega \mathrm{t}$ $\Rightarrow \mathrm{V}=\mathrm{A} \omega \cos \omega \mathrm{t}$ Now, $\begin{aligned} & \mathrm{V}_{	ext {mean }}=\frac{\int_0^{\mathrm{T} / 12} \mathrm{~A} \omega \cos \omega \mathrm{dt}}{\int_0^{\mathrm{T} / 12} \mathrm{dt}}=\frac{\frac{\mathrm{A} \omega}{\omega}[\sin \omega \mathrm{t}]_0^{\mathrm{T} / 12}}{\mathrm{~T} / 12} \ & =\frac{\mathrm{A}\left[\frac{1}{2}-0ight]}{\frac{\mathrm{T}}{12}}=\frac{\frac{0.1}{2}}{\frac{0.6}{12}}=\frac{12}{12}=1 \mathrm{~m} / \mathrm{s} \end{aligned}$

A particle executes simple harmonic motion with a time period $0.6 \mathrm{~s}$ and amplitude $10…

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