Question

A particle executes simple harmonic motion with an angular velocity of 3.5 m sec^-1 and the maximum acceleration 7.5 m sec^-2.Find the amplitude of oscillation?​

Answer 1

Explanation:

ω=3.5 radian/sec

Maximum acceleration of a particle under SHM=ω

2

a where a is amplitude of oscillation.

ω

2

a=7.5⇒(3.5)

2

a=7.5

⇒a=

3.5×3.5

7.5

⇒=

49

30

=0.61m