A simple pendulum of length 1m is oscillated at a place where g = 9.8m/s². Find the maximum velocity of the bob of the simple pendulum if the amplitude of oscillation is 3cm.

Complete Solution: 

we calculate the maximum velocity (v_max) of the bob of a simple pendulum using the formula: v_max = ω · A

The angular frequency is given by: ω = √(g / L)

 Here:

g = 9.8 m/s²

L = 1 m

Substitute the values:

ω = √(9.8 / 1) = √9.8 ≈ 3.13 rad/s

The amplitude is given as 3 cm, so:

A = 3 / 100 = 0.03 m

Using the formula:

v_max = ω · A

Substitute ω = 3.13 rad/s and A = 0.03 m:

v_max = 3.13 · 0.03 ≈ 0.0939 m/s

Convert v_max to cm/s:

v_max = 0.0939 · 100 = 9.391 cm/s

Final Answer:

The maximum velocity of the bob is 9.391cms⁻¹