Q119 / 200 · NEET 2007 · Physics

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
    $T / 8$
  • Correct answer
    B
    $T / 12$
  • C
    $T / 2$
  • D
    $T / 4$.

Solution

For equilibrium position $\begin{aligned} x(t) & =a \sin \omega t \\ \text {At } x(t) & =\frac{a}{2} \\ \therefore \sin \left(\frac{\pi}{6}\right) & =\sin \omega t\left\{\because \omega=\frac{2 \pi}{\mathrm{T}}\right\} \\ \text {or } \frac{\pi}{6} & =\frac{2 \pi t}{T} \\ \Rightarrow t & =\frac{T}{12} \end{aligned}$

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

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