NEET | 2017SHMHard

Question

A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is :-

Options

A.

52π

B.

4π5

C.

2π3

D.

5π

Solution

Amplitude A = 3 cm
When particle is at x = 2 cm ,
its |velocity| = |acceleration|
i.e., ωA2-x2=ω2x ω = A2-x2x
T = 2πω=2π25=4π5

Create a free account to view solution

View Solution Free
Topic: SHM·Practice all SHM questions

More SHM Questions