A particle performing $SHM$ is found at its equilibrium at $t = 1\,sec$. and it is found to have a speed of $0.25\,m/s$ at $t = 2\,sec$. If the period of oscillation is $6\,sec$. Calculate amplitude of oscillation
Diffcult
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$x=A \sin (\omega t+\phi)$
$\mathrm{O}=\mathrm{A} \sin (\omega+\phi)$ as $\mathrm{x}=0$ at $\mathrm{t}=1 \mathrm{s}$
and $\mathrm{v}=\mathrm{A} \omega \cos (\omega \mathrm{t}+\phi)$
$\frac{1}{4}=\mathrm{A} \omega \cos (2 \omega+\phi)$ as $\mathrm{v}=\frac{1}{4} \frac{\mathrm{M}}{\mathrm{s}}$ at $\mathrm{t}=2 \mathrm{s}$
$\text { here } \omega=\frac{2 \pi}{6}=\frac{\pi}{3} \quad \text { as } \mathrm{T}=6 \mathrm{s}$
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