Solveeit Logo
Login

Question

Question: A particle executing simple harmonic motion with amplitude of 0.1 m. At a certain instant when its d...

A particle executing simple harmonic motion with amplitude of 0.1 m. At a certain instant when its displacement is 0.02 m, its acceleration is 0.5 m/s2. The maximum velocity of the particle is (in m/s)

A

0.01

B

0.05

C

0.5

D

0.25

Answer

0.5

Explanation

Solution

Acceleration A=ω2y⇒ω=Ay=0.50.02=5A = \omega^{2}y \Rightarrow \omega = \sqrt{\frac{A}{y}} = \sqrt{\frac{0.5}{0.02}} = 5

Maximum velocity

vmax⁡v_{\max}