Question 24 of 194

A particle executes simple harmonic motion between $x=–A\text{and}x=+A$. If time taken by particle to go from $x=0$ to $\dfrac{A}{2}$ is $2s$; then time taken by particle in going from $x=\dfrac{A}{2}$ to $A$ is:

A$3s$B$2s$C$1.5s$D$4s$

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Answer: (D) $4s$

The position of a particle performing an SHM is given by:
$x=A\sin (\omega t+\phi )$
Since at $t=0s$, the particle is at $x=0$, therefore the initial phase $\phi =0$.
At time $t=2s$, the particle is at $x=\dfrac{A}{2}$.
$∴\dfrac{A}{2}=A\sin \left(2\omega \right) \Rightarrow \omega =\dfrac{\pi }{12}{s}^{-1}$
Therefore, time period
$T=\dfrac{2\pi }{\omega }=24s$
The total time taken from $x=0$ to $x=A$ is $\dfrac{T}{4}=6s$.
Since the time taken from $x=0$ to $x=\dfrac{A}{2}$ is $2s$, therefore the time taken from $x=\dfrac{A}{2}$ to $x=A$ is $4s$.