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

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 (A)  (T/4)  (B)  (T/8)  (C)  (T/12)  (D)  (T/2)

Tardigrade Admin · Accepted Answer

Let displacement equation of particle executing, SHM is

y = a sin ω t

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

y = \frac{a}{2}

Therefore,

\frac{a}{2} = a sin ω t

or

sin ω t = \frac{1}{2} = sin \frac{π}{6}

or

ω t = \frac{π}{6}

or

t = \frac{π}{6 ω}

or

t = \frac{π}{6 ( \frac{2 π}{T} )}

(a s 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 .

Q. 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

$\frac{T}{4}$

$\frac{T}{8}$

$\frac{T}{12}$

$\frac{T}{2}$

Q. 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

4T​

8T​

12T​

2T​

$\frac{T}{4}$

$\frac{T}{8}$

$\frac{T}{12}$

$\frac{T}{2}$