The equation of motion of a particle performing linear S.H.M is $x=5 \sin \left[4 t-\frac{\pi}{6}\right]$,…

The equation of motion of a particle performing linear S.H.M is $x=5 \sin \left[4 t-\frac{\pi}{6}\right]$, where $x$ is its displacement in $\mathrm{cm}$. The velocity of the particle when its displacement is $3 \mathrm{~cm}$, is
  1. $8 \mathrm{~cm} / \mathrm{s}$
  2. $6 \mathrm{~cm} / \mathrm{s}$
  3. $16 \mathrm{~cm} / \mathrm{s}$
  4. $10 \mathrm{~cm} / \mathrm{s}$

Solution

Comparing with standard equation: $x=a \sin [\omega t+\phi]$ The amplitude of the wave is $a=5$ and the angular velocity of the wave is $\omega=4$ The velocity of the wave can be obtained by taking derivative as: $\begin{aligned} & \frac{d x}{d t}=\omega a \cos [\omega t+\phi]=\omega \sqrt{a^2-x^2} \\ & \therefore v=\omega \sqrt{a^2-y^2}=4 \sqrt{(5)^2-(3)^2}=16 \mathrm{~cm} / \mathrm{s}\end{aligned}$ .

Asked in: MHT CET 2022 (06 Aug Shift 2)

Practice more Oscillations questions on Aicharya