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If $x=5 \sin \left(\pi t+\frac{\pi}{3}\right) \mathrm{m}$ represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are
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$5 \mathrm{~m}, 2 \mathrm{~s}$
$\begin{aligned} & \therefore x=5 \sin \left(\pi t+\frac{\pi}{3}\right) \mathrm{m} \\ & \text { Amplitude }=5 \mathrm{~m} \\ & \omega=\pi=\frac{2 \pi}{T} \\ & T=\frac{2 \pi}{\pi}=2 \mathrm{~s}\end{aligned}$
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