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In a science museum, a 110 kg brass pendulum bob swings at the end of a 15.0-m-long wire. The pendulum is started at exactly 8:00 a.m. every morning by pulling it 1.5 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?
ANSWER:
The damping time constant is
m/b=11000s.n t2 4 x 3600 [tex] \sqrt{9.8/ 15 - 0.01 x^{2}/ 4 x 110 x^{2} /2 \pi [/tex] = 1852.5
In a science museum, a 110 kg brass pendulum bob swings at the end of a 15.0-m-long wire. The pendulum is started at exactly 8:00 a.m. every morning by pulling it 1.5 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?
ANSWER:
The damping time constant is
m/b=11000s.n t2 4 x 3600 [tex] \sqrt{9.8/ 15 - 0.01 x^{2}/ 4 x 110 x^{2} /2 \pi [/tex] = 1852.5
The pendulum will complete [tex]\boxed{1872}[/tex] oscillations between [tex]8:00\,{\text{am}}[/tex] to [tex]12:00\,{\text{pm}}[/tex] .
Further Explanation:
Given:
The length of the simple pendulum is [tex]15\,{\text{m}}[/tex] .
The pendulum is started at [tex]8:00\,{\text{am}}[/tex] .
The pendulum is stopped at [tex]12:00\,{\text{pm}}[/tex] .
Concept:
The frequency of oscillation of the simple pendulum is given by:
[tex]f=\frac{1}{{2\pi}}\sqrt{\frac{g}{l}}[/tex]
(1)
Here, [tex]f[/tex] is the frequency of oscillation of pendulum, [tex]g[/tex] is the acceleration due to gravity and [tex]l[/tex] is the length of the pendulum.
The value of acceleration due to gravity on the surface of Earth is [tex]9.8\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}[/tex]
Substitute the values in equation (1).
[tex]\begin{aligned}f&=\frac{1}{{2\pi}}\sqrt{\frac{{9.8}}{{15}}}\\&=\frac{1}{{2\pi}}\left({0.808}\right)\\&\approx0.13\,{\text{Hz}}\\\end{aligned}[/tex]
It means that pendulum makes [tex]0.13[/tex] oscillation in [tex]1\,\sec[/tex] .
The number of seconds from [tex]8:00\,{\text{am}}[/tex] to [tex]12:00\,{\text{pm}}[/tex] is:
[tex]\begin{aligned}t&=\left({12 - 8}\right)\left({3600\,}\right)\,{\text{s}}\\&=4\times3600\,{\text{s}}\\&=14400\,{\text{s}}\\\end{aligned}[/tex]
So number of oscillations completed by the pendulum is:
[tex]\begin{aligned}n7&=f\times t\\&=0.13\times14400\\&=1872\,{\text{oscillations}}\\\end{aligned}[/tex]
Thus, the pendulum will complete [tex]\boxed{1872}[/tex] oscillations from [tex]8:00\,{\text{am}}[/tex] to [tex]12:00\,{\text{pm}}[/tex] .
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Answer details:
Grade: High School
Subject: Physics
Chapter: Simple Pendulum
Keywords:
Pendulum, oscillations, 8:00 am, 12:00 pm, length of pendulum 15m, 110kg, brass pendulum, bob swings, frequency of pendulum, number of oscillations.
