A body executing $S.H.M.$ along a straightline has a velocity of $3 \,ms ^{-1}$ when it is at a distance of $4 \,m$ from its mean position and $4 \,ms ^{-1}$ when it is at a distance of $3 \,m$ from its mean position. Its angular frequency and amplitude are
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(d)
$v=\omega \sqrt{A^2-x^2}$
$v_1=3 \,m / s x_1=4 \,m$
$v_2=4 \,m / s x_2=3 \,m$
$3=\omega \sqrt{A^2-4^2} \ldots(i)$
$4=\omega \sqrt{A^2-3^2} \ldots(ii)$
Solving $(i)$ and $(ii)$, we get
$A=5 \,m$ and $\omega=1 \,rad / s$
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