First slide

Simple hormonic motion

Question

A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is:

Moderate

A

$\frac{T}{4}$
B

$\frac{T}{8}$
C

$\frac{T}{12}$
D

$\frac{T}{2}$

Solution

Let displacement equation of particle executing SHM is $y = a \sin ωt$
As particle travels half of the amplitude from the equilibrium position, so y = $\frac{a}{2}$
Therefore, $\frac{a}{2} = a \sin ωt$
or $\sin ωt = \frac{1}{2} = \sin \frac{π}{6} or ωt = \frac{π}{6}$
or $t = \frac{π}{6 ω} or t = \frac{π}{6 (\frac{2 π}{T})} (as ω = \frac{2 π}{T})$
or t = $\frac{T}{12}$
Hence, the particle travels half of the amplitude from the equilibrium in $\frac{T}{12} \sec$

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