Question

Part A In a science museum, a 120 kg brass pendulum bob swings at the end of a 13.9 m-long wire. The pendulum is started at exactly 8:00 a.m. every morning by pulling it 1.6 m to the side and releasing it. Because of its compact shape and smooth surface, the pendulum's damping constant is only 0.010 kg/s. At exactly 12:00 noon, how many oscillations will the pendulum have completed? Express your answer as an integer. You may want to review (Pages 405 - 407). IVO AQ R O 2 ? N= Submit Request Answer Part B What is its amplitude at noon? Express your answer to two significant figures and include the appropriate units. ? T HÅR Value O 2 Units

Answer 1

A) The frequency of oscillation of the simple pendulum is given by:

f = 1/2π * √(g/l) = 1/2π * √(9.8/13.9) => f = 0.133 Hz

Here, f is the frequency of oscillation of pendulum, g is the acceleration due to gravity and l is the length of the pendulum.

It means that pendulum makes oscillation in $1\,\sec$ .

The number of seconds from $8:00\,{\text{am}}$ to $12:00\,{\text{pm}}$ is:

(12 - 8) (3600 t =4 x 3600s = 14400 s

So number of oscillations completed by the pendulum is:

n = f * t = 0.133*14400 = 1924 oscillations

Thus, the pendulum will complete 1924 oscillations from $8:00\,{\text{am}}$  to $12:00\,{\text{pm}}$ .

B)

The amplitude at noon will be 1.01 meters.
1.6 exp( -0.01 kg/s * 4 hours / (2*120 kg)) = 1.01m