Question

The acceleration of a particle executing simple
harmonic motion is (π²/2) cm s^-2 .when its
displacement is 2 cm. Compute its time period.
Ans. 4 s.
​

Answer 1

Answer:

Given:

Acceleration of a particle undergoing SHM is π²/2 when displacement is 2cm.

To find:

Time period of the SHM.

Formulas used:

1. Acceleration = - ω²x

2. Time period = (2π/ω).

Calculation:

As per the question:

Acceleration = - π²/2

=> - ω²x = - π²/2

=> ω²2 = π²/2

=> ω² = π²/4

=> ω = π/2 .......(1)

Now, time period is as follows:

T = 2π/ω

=> T = 2π/(π/2)

=> T = 4 seconds

So final answer is 4 seconds.