A particle is executing simple harmonic motion. The magnitude of its velocity at the mean position is 10 cm/s and the magnitude of its acceleration at the extreme position is 50 cm/s ^2 . If the time period of the motion is x 5 s , then the value of x is _______.

Step-by-step solution

The magnitude of velocity at the mean position is the maximum velocity of the particle. v_ = A = 10 cm/s The magnitude of acceleration at the extreme position is the maximum acceleration of the particle. a_ = ^2 A = 50 cm/s ^2 Dividing the maximum acceleration by the maximum velocity gives the angular frequency : = ^2 A A = 50 10 = 5 rad/s The time period T of the simple harmonic motion is: T = 2 = 2 5 s Comparing this with the given form x 5 s , we find x = 2 . Answer: 2