Q134 / 180 · KCET 2016 · Physics

A particle executing SHM has a maximum speed of \( 0.5 \mathrm{~ms}^{-1} \) and maximum acceleration of
\( 1.0 \mathrm{~ms}^{-2} \). The angular frequency of oscillation is

  • Correct answer
    A
    \( 2 \operatorname{rad} \mathrm{s}^{-1} \)
  • B
    \( 0.5 \operatorname{rad} \mathrm{s}^{-1} \)
  • C
    \( 2 \Pi \mathrm{rad} \mathrm{s}^{-1} \)
  • D
    \( 0.5 \Pi \mathrm{rad} \mathrm{s}^{-1} \)

Solution

Given, maximum speed, $v_{\max }=0.5 m s^{-1} ;$ maximum acceleration,$a_{\max }=1.0 m s^{-1} ;$ angular frequency, $\omega=?$ Now, we know $v_{\max }=\omega A \rightarrow(1)$ where A is amplitude of oscillation. Also, $a_{\max }=\omega^{2} A \rightarrow(2)$ Using Eqs. (1) and (2), we get $\frac{v_{\max }}{a_{\max }}=\frac{\omega A}{\omega^{2} A}$ $\Rightarrow \frac{v_{\max }}{a_{\max }}=\frac{1}{\omega}$ $\Rightarrow \omega=\frac{a_{\max }}{v_{\max }}=\frac{1.0}{0.5}$ $\Rightarrow \omega=2 \operatorname{rad} s^{-1}$ Therefore, angular frequency of oscillation is $2 \mathrm{rads}^{-1}$

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

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