A particle executing simple harmonic motion of amplitude $5 \mathrm{~cm}$ has maximum speed of $31.4…

A particle executing simple harmonic motion of amplitude $5 \mathrm{~cm}$ has maximum speed of $31.4 \mathrm{~cm} / \mathrm{s}$. The frequency of its oscillation is:
  1. $4 \mathrm{~Hz}$
  2. $3 \mathrm{~Hz}$
  3. $2 \mathrm{~Hz}$
  4. $1 \mathrm{~Hz}$

Solution

Here $a=5 \mathrm{~cm}, V_{\max }=\frac{31.4 \mathrm{~cm}}{\mathrm{~s}}$ $\begin{aligned} & V_{\max }=\omega a=31.4=2 \pi \mathrm{v} \times 5 \\ & \Rightarrow 31.4=10 \times 3.14 \times V \\ & \Rightarrow \quad V=1 Hz \end{aligned}$ :

Asked in: NEET 2005

Practice more Oscillations questions on Aicharya