Question

The length of a simple pendulum is 0.79m and the mass of the particle at the end of the cable is 0.24 kg. The pendulum is pulled away from its equilibrium position by an angle of 8.5 degrees and released from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion. What is the angular frequency of the motion? What is bobs speed as it passes through the lowest point of the spring?

STEPS PLEASE. answer doesnt matter.

Answer 1

The angular frequency of the harmonic motion is
w = sqrt (g/L) radians per second,
and is independent of mass. Calculate it.

This is an accurate approximate formula that is almost always used. The approximation
sin theta = theta
is made in the derivation.
There is a small correction for angles greater than about 15 degrees.

The bob's maximum horizontal deflection from vertical is
A = L sin 8.5 = 0.117 m

The bob's x position is
x = A cos wt

The maximum velocity (at the lowest point) is w *A