Question
A particle is executing SHM of periodic time $\mathrm{T}$. The time taken by a particle in moving from mean position to half the maximum displacement is $\left(\sin 30^{\circ}=\right.$ $0.5):$(a) $\mathrm{T} / 2$(b) $\mathrm{T} / 4$(c) $\mathrm{T} / 8$(d) $\mathrm{T} / 12$
Step 1
Step 1: We know that the displacement in simple harmonic motion is given by $x = A \sin(\omega t)$, where $A$ is the amplitude, $\omega$ is the angular frequency, and $t$ is the time. Show more…
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Round 2
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