Question

2. A pendulum exhibits simple harmonic motion when it is released at an angle θ = 9°. The length of the pendulum is l=0.86 m, and the pendulum bob has a mass m=0.3 kg.

a. What is the pendulum’s frequency?

b. Use energy conservation to find the pendulum bob’s speed when it passes through the lowest point of the swing? Assume that the difference in height between the lowest part of the swing and the highest part of the swing is delta H= l(1-cos(θ))

c. What is the total energy of the system?

Answer 1

a) Since the value of theta is very small, we consider it as a small amplitude oscillation.

Therefore, the time period of a simple Pendulum is given by

T=2π√(l/g)

Where l is the length of the pendulum and g is the acceleration due to gravity

l=0.86m

g=9.8m/s^2

Substituting the values in the above equation,

T=1.86s

Frequency=1/ T

=1/1.86=0.537 Hz

b) using energy conservation,

The value of kinetic energy is maximum at the lowest point of the pendulum and there, the potential energy is minimun, that is zero. And the potential energy has maximum value at the highest point of the pendulum, there is the value of kinetic energy is minimum. That is the total mechanical energy is conserved.

Maximum kinetic energy=maximum potential energy

So we can write (1/2) m×v^2=mgh

Where m is the mass v is the velocity, g is the acceleration due to gravity and he is the height difference.

Therefore v^2= 2gh

Given h=l(1-cos (theta))