In a science museum, a 110 kg brass pendulum bob swings at the end of a 13.2 m -long wire. The pendulum is started at exactly 8:00 a.m. every morning by pulling it 1.7 m to the side and releasing it. Because of its compact shape and smooth surface, the pendulum's damping constant is only 0.010kg/s.
A. At exactly 12:00 noon, how many oscillations will the pendulum have completed?
B. And what is its amplitude?
Answer
(A) The expression of time period of the pendulum is,
T=2 \pi \sqrt{\frac{l}{g}}
The expression of the frequency is,
f=\frac{1}{T}
Substitute 2 \pi \sqrt{\frac{l}{g}} for T.
f=\frac{1}{2 \pi} \sqrt{\frac{g}{l}}
Substitute 13.2 \mathrm{~m} for \mathrm{l} and 9.8 \mathrm{~m} / \mathrm{s}^{2} for \mathrm{g} to find frequency.
f=\frac{1}{2 \pi} \sqrt{\frac{9.8 \mathrm{~m} / \mathrm{s}^{2}}{13.2 \mathrm{~m}}}
=0.137 oscillation/ \mathrm{s}
The time interval between 8:00 AM to 12.00 PM is,
t=4 \mathrm{~h}
The number of oscillations completed in time t is, N=f t
Substitute 0.137 oscillation/s for f and 4 \mathrm{~h} for t.
N=(0.137 oscillation/ \mathrm{s})\left(4 \mathrm{~h}\left(\frac{3600 \mathrm{~s}}{1 \mathrm{~h}}\right)\right)
=1973 oscillation
The number of oscillations of the pendulum is 1973
(B) The expression of the amplitude at time t is,
x(t)=A \exp \left(-\frac{b t}{2 m}\right)
Substitute 110 \mathrm{~kg} for \mathrm{m}, 1.7 \mathrm{~m} for \mathrm{A}, 0.010 \mathrm{~kg} / \mathrm{s} for \mathrm{b}, and 4 \mathrm{~h} for \mathrm{t}.
x(t) =(1.7 \mathrm{~m}) \exp \left(-\frac{(0.010 \mathrm{~kg} / \mathrm{s})\left(4 \mathrm{~h}\left(\frac{3600 \mathrm{~s}}{1 \mathrm{~h}}\right)\right)}{2(110 \mathrm{~kg})}\right)
=0.883 \mathrm{~m}\left(\frac{10^{-2} \mathrm{~m}}{1 \mathrm{~m}}\right)
=88.3 \mathrm{~cm}
The amplitude is 88.3 \mathrm{~cm}.
