Q162 / 200 · Manipal (MET) 2011 · Physics

The acceleration of a particle performing SHM is $12 \mathrm{~cm} / \mathrm{s}^2$ at a distance of 3 cm from the mean position. Its time period is

  • A
    0.5 s
  • B
    1.0 s
  • C
    2.0 s
  • Correct answer
    D
    3.14 s

Solution

Acceleration $a=\omega^2 y$ $\begin{aligned} \Rightarrow \quad a & =\left(\frac{2 \pi}{T}\right)^2 y \\ \Rightarrow \quad T^2 & =\frac{4 \pi^2 y}{a} \\ \Rightarrow \quad T & =2 \pi \sqrt{\frac{y}{a}} \\ & =2 \pi \sqrt{\frac{3}{12}} \\ T & =2 \pi \times \frac{1}{2}=3.14 \mathrm{~s}\end{aligned}$ $T=2 \pi \times \frac{1}{2}=3.14 \mathrm{~s}$

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

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