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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Aicharya · Accepted Answer

$\begin{aligned} & 	herefore x=5 \sin \left(\pi t+\frac{\pi}{3}ight) \mathrm{m} \ & 	ext { Amplitude }=5 \mathrm{~m} \ & \omega=\pi=\frac{2 \pi}{T} \ & T=\frac{2 \pi}{\pi}=2 \mathrm{~s}\end{aligned}$ .

If $x=5 \sin \left(\pi t+\frac{\pi}{3}\right) \mathrm{m}$ represents the motion of a particle executing…

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