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A particle performs simple harmonic motion with a time period of $16 \mathrm{~s}$. At a time $t=2 \mathrm{~s}$, the particle passes through the origin and at $\mathrm{t}=4 \mathrm{~s}$ its velocity is $4 \mathrm{~m} / \mathrm{s}$. The amplitude of the motion is
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The correct answer is:
$\frac{32 \sqrt{2}}{\pi}$
We have
$\begin{aligned} & \Delta \phi==\frac{2 \pi}{\mathrm{T}} \times \mathrm{t}=\frac{2 \pi}{16} \times 2=\frac{\pi}{4} \\ & \text { So, } \mathrm{V}=\mathrm{A} \omega \cos (\omega \mathrm{t}+\pi / 4) \\ & 4=\mathrm{A} \omega \cos \left(\frac{2 \pi}{16} \times 4+\pi / 4\right) \\ & \Rightarrow \quad 4=\mathrm{A} \omega \sin \pi / 4 \Rightarrow 4=\mathrm{A} \times \frac{2 \pi}{16} \times \frac{1}{\sqrt{2}} \\ & \Rightarrow \quad \mathrm{A}=\frac{64 \sqrt{2}}{2 \pi} \Rightarrow \mathrm{A}=\frac{32 \sqrt{2}}{\pi}\end{aligned}$
$\begin{aligned} & \Delta \phi==\frac{2 \pi}{\mathrm{T}} \times \mathrm{t}=\frac{2 \pi}{16} \times 2=\frac{\pi}{4} \\ & \text { So, } \mathrm{V}=\mathrm{A} \omega \cos (\omega \mathrm{t}+\pi / 4) \\ & 4=\mathrm{A} \omega \cos \left(\frac{2 \pi}{16} \times 4+\pi / 4\right) \\ & \Rightarrow \quad 4=\mathrm{A} \omega \sin \pi / 4 \Rightarrow 4=\mathrm{A} \times \frac{2 \pi}{16} \times \frac{1}{\sqrt{2}} \\ & \Rightarrow \quad \mathrm{A}=\frac{64 \sqrt{2}}{2 \pi} \Rightarrow \mathrm{A}=\frac{32 \sqrt{2}}{\pi}\end{aligned}$
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