Question
The maximum velocity and maximum acceleration of a particle executing S.H.M. are $1 \mathrm{~m} / \mathrm{s}$ and $3.14 \mathrm{~m} / \mathrm{s}^{2}$ respectively. The frequency of oscillation for this particle is......(A) $0.5 \mathrm{~s}^{-1}$(B) $3.14 \mathrm{~s}^{-1}$(C) $0.25 \mathrm{~s}^{-1}$(D) $2 \mathrm{~s}^{-1}$
Step 1
14 m/s$^2$. We know that $V_m = \omega A$ and $A_m = \omega^2 A$ for a particle undergoing simple harmonic motion (SHM), where $\omega$ is the angular frequency and $A$ is the amplitude of the motion. Show more…
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