The maximum velocity of a particle performing simple harmonic motion is $6.28 \mathrm{~cm} \mathrm{~s}^{-1}$…

The maximum velocity of a particle performing simple harmonic motion is $6.28 \mathrm{~cm} \mathrm{~s}^{-1}$. If the length of its path is $8 \mathrm{~cm}$, then what is its period?
  1. $2 \mathrm{~s}$
  2. $4 \mathrm{~s}$
  3. $3 \mathrm{~s}$
  4. $1 \mathrm{~s}$

Solution

In SHM, $ v_{\max }=6.28 \mathrm{~cm} \mathrm{~s}^{-1} $ Length of path $=8 \mathrm{~cm}$ $\therefore$ Amplitude of particle, $a=\frac{\text { Length of path }}{2}$ $ =\frac{8}{2}=4 \mathrm{~cm} $ $ \begin{aligned} & \therefore & v_{\text {max }} & =\omega a \\ \Rightarrow & & \omega & =v_{\text {max }} / a \\ \Rightarrow & & \omega & =\frac{6.28}{4} \mathrm{rad} \mathrm{s}^{-1} \\ \Rightarrow & & \frac{2 \pi}{T} & =\frac{6.28}{4} \\ \Rightarrow & & T & =\frac{4 \times 2 \pi}{6.28}=\frac{4 \times 2 \times 3.14}{6.28}=4 \mathrm{~s} \end{aligned} $

Asked in: AP EAMCET 2020 (22 Sep Shift 2)

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