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

$\frac{T}{4}$

$\frac{T}{8}$

$\frac{T}{12}$

$\frac{T}{2}$

Solution

$\frac{T}{12}$

Let displacement equation of particle executing SHM is

$y = α sinωt$

As the particle travels half of the amplitude from the equilibrium position, so

$y = \frac{α}{2} ∴ \frac{α}{2} = α sinωt \Rightarrow \sin ωt = \frac{1}{2} = \sin \frac{π}{6} \Rightarrow ωt = \frac{1}{2} = \sin \frac{π}{6} \Rightarrow ωt = \frac{π}{6} \Rightarrow t = \frac{π}{6 ω}$

$\Rightarrow t = \frac{π}{6 (\frac{2 π}{T})} \Rightarrow t = \frac{T}{12}$

Hence, the particle travels half of the amplitude from the equilibrium in $\frac{T}{12} s$ .

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

$\frac{T}{4}$

$\frac{T}{8}$

$\frac{T}{12}$

$\frac{T}{2}$

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For Daily Free Study Material Join wiredfaculty

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A particle executes simple harmonic oscillation with an amplitude α . The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is T4 T8 T12 T2

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A particle of mass m executes simple harmonic motion with amplitude ‘a’ and frequency ‘ν’. The average kinetic energy during its motion from the position of equilibrium to the end is

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Entrance Exams Preparation

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

$\frac{T}{4}$

$\frac{T}{8}$

$\frac{T}{12}$

$\frac{T}{2}$

The function sin²(ωt) represents