Q43 / 214 · NTA Abhyas (NEET) 2020 · Physics · Oscillations

A particle executes a simple harmonic motion of time period $T$ . Find the time taken by the particle to go directly from its mean position to half the amplitude.

A
$\frac{T}{2}$
B
$\frac{T}{4}$
C
$\frac{T}{8}$
Correct answer
D
$\frac{T}{12}$

Solution

y = A \sin ω t = A \sin (\frac{2 π}{T} t)

\Rightarrow \frac{A}{2} = A \sin (\frac{2 π t}{T})

\Rightarrow t = \frac{T}{12}

Hence, the correct answer is option: 'D'.

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