Q143 / 180 · COMEDK 2022 · Physics
A particle is performing simple harmonic motion. Equation of its motion is $x=5 \sin \left(4 t-\dfrac{\pi}{6}\right), x$ being the displacement from mean position. Velocity (in $\mathrm{ms}^{-1}$ ) of the particle at the instant when its displacement is 3 , will be
- A$\dfrac{2 \pi}{3}$
- B$\dfrac{5 \pi}{6}$
- C20
- Correct answerD16
Solution
The displacement equation is given by $x = 5 \sin \left(4t - \dfrac{\pi}{6}\right)$.
The velocity $v$ of a particle in simple harmonic motion is given by $v = \dfrac{dx}{dt} = 5 \times 4 \cos \left(4t - \dfrac{\pi}{6}\right) = 20 \cos \left(4t - \dfrac{\pi}{6}\right)$.
Using the identity $\cos \theta = \sqrt{1 - \sin^2 \theta}$, we have $v = 20 \sqrt{1 - \sin^2 \left(4t - \dfrac{\pi}{6}\right)}$.
Given the displacement $x = 5 \sin \left(4t - \dfrac{\pi}{6}\right) = 3$, we find $\sin \left(4t - \dfrac{\pi}{6}\right) = \dfrac{3}{5}$.
Substituting this value into the velocity expression:
$v = 20 \sqrt{1 - \left(\dfrac{3}{5}\right)^2} = 20 \sqrt{1 - \dfrac{9}{25}} = 20 \sqrt{\dfrac{16}{25}} = 20 \times \dfrac{4}{5} = 16$.
Answer: 16
The velocity $v$ of a particle in simple harmonic motion is given by $v = \dfrac{dx}{dt} = 5 \times 4 \cos \left(4t - \dfrac{\pi}{6}\right) = 20 \cos \left(4t - \dfrac{\pi}{6}\right)$.
Using the identity $\cos \theta = \sqrt{1 - \sin^2 \theta}$, we have $v = 20 \sqrt{1 - \sin^2 \left(4t - \dfrac{\pi}{6}\right)}$.
Given the displacement $x = 5 \sin \left(4t - \dfrac{\pi}{6}\right) = 3$, we find $\sin \left(4t - \dfrac{\pi}{6}\right) = \dfrac{3}{5}$.
Substituting this value into the velocity expression:
$v = 20 \sqrt{1 - \left(\dfrac{3}{5}\right)^2} = 20 \sqrt{1 - \dfrac{9}{25}} = 20 \sqrt{\dfrac{16}{25}} = 20 \times \dfrac{4}{5} = 16$.
Answer: 16
Hence, the correct answer is option: 'D'.