Question

An object executing SHM has its maximum velocity
0.16m/s. If it goes 0.02m either side of mean position,
find its time period and maximum acceleration.

Answer 1

Given :

Max. linear velocity = 0.16m/s

Amplitude = 0.02m

To Find :

Time period and maximum linear acceleration of the object$.$

Solution :

❒ First of all we need to find angular velocity of the object.

Maximum velocity of particle executing SHM is given by, v = A ω

• v denotes linear velocity
• A denotes amplitude
• ω denotes angular velocity

By substituting the given values;

➙ v = A ω

➙ 0.16 = 0.02 × ω

➙ ω = 0.16/0.02

➙ ω = 8 rad/s

A] Time period of parties :

Relation between time period of particle executing SHM and angular velocity is given by, T = 2π / ω

➙ T = 2π / ω

➙ T = (2 × 3.14) / 8

➙ T = 3.14/4

➙ T = 0.785 s

B] Max. linear acceleration :

Formula : a = -ω² A

➙ a = -ω² A

➙ a = -8² × 0.02

➙ a = -64 × 0.02

➙ a = -1.28 m/s²

[Negative sign shows that acceleration is directed towards mean position.]