Question
A particle executing S.H.M. has an amplitude $\mathrm{A}$ and periodic time $\mathrm{T}$. The minimum time required by the particle to get displaced by $(\mathrm{A} / \sqrt{2})$ from its equilibrium position is $\ldots \ldots \ldots \mathrm{s}$.(A) $\mathrm{T}$(B) $\mathrm{T} / 4$(C) $\mathrm{T} / 8$(D) $\mathrm{T} / 16$
Step 1
H.M) with amplitude A and periodic time T. The displacement of the particle from its equilibrium position is given by the equation of S.H.M, which is $x = A \sin(\omega t)$, where $\omega$ is the angular frequency and $t$ is the time. Show more…
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