Question

4. A Foucault pendulum is used to demonstrate the rotation of the Earth. They can usually be found in science museums. A certain pendulum has a 120-kg bob swinging on a 16.2-m-long wire with a maximum amplitude of 6.3 A. Sketch the pendulum at its maximum displacement and label all known quantities. B. What is the difference in gravitational potential energy between the maximum displacement and the equilibrium point? C. What is the maximum speed of the bob? D. Where in the oscillation is the bob when it has its maximum speed? E. What is the period and frequency of the oscillation?

Answer 1

Hi,

Hope you are doing well.

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PART A:

Lcose L-Lcose

$\theta$ is the amplitude of oscillation, is the length of the wire.

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PART B:

From figure, we have, the maximum height reached by the pendulum is given by:

$h=L-L\cos\theta$

Given that,

Mass of the bob,

m 120 kg

Length of the wire, $L=16.2\;m$

Acceleration due to gravity,

g9.8 m/s

Therefore, Gravitational potential energy (assuming that the lowest position of the bob is at zero height) is given by:

$\Delta U=mgh=mg(L-L\cos\theta)=mgL(1-\cos\theta)$

$\therefore \Delta U=120\;kg\times16.2\;m\times9.8\;m/s^{2}(1-\cos6.3^{o})$

$\boldsymbol{\therefore \Delta U=115.05\;J}$

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PART C:

The maximum speed of the bob will be at equilibrium position or at the lowest point.

According the Law of conservation of energy, Total mechanical energy at the maximum height (only Potential energy) will be converted into total mechanical energy at the lowest point (only kinetic energy).

$\therefore \Delta U=K.E$

$\therefore \Delta U=\frac{1}{2}mv^{2}$

Therefore, Maximum speed of the bob is given by,

$\therefore V=\sqrt{ \frac{2\Delta U}{m}}=\sqrt{\frac{2\times115.05\;J}{120\;kg}}$

$\therefore V=\sqrt{ 1.9175}\;m/s$

$\boldsymbol{\therefore V=1.385\;m/s}$

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PART D:

The bob will be at equilibrium position or at the lowest point when it has maximum speed because the potential energy is completely converted into kinetic energy at this point, since at the lowest point potential energy is zero.

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PART E:

Period of the oscillation is given by:

$T=2\pi\sqrt{\frac{L}{g}}=2\pi\sqrt{\frac{16.2\;m}{9.8\;m/s^{2}}}$

$\therefore T=2\pi\times 1.2857\;s$

$\boldsymbol{\therefore T=8.08\;s}$

Frequency of the oscillation is given by:

$f= \frac{1}{T}=\frac{1}{8.08\;s}$

$\boldsymbol{\therefore f=0.124\;s^{-1}=0.124\;Hz}$

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