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 second is
- 52
- 4 5
- 2 3
- 5
Answer
(2) 4 5
Solution
In SHM, v = A2 - x2 and |a| = 2 x
|v| = |a| at x = 2 cm, A = 3 cm
32 - 22 = 2 2
= 52 rad s-1
T = 2 = 4 5 s
Answer: (2)