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Physics Principles with Applications
Found in: Page 138
Physics Principles with Applications

Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922

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Short Answer

The length of a simple pendulum is 0.72 m, the pendulum bob has a mass of 295 g, and it is released at an angle of \(12^\circ \) to the vertical. Assume SHM. (a) With what frequency does it oscillate? (b) What is the pendulum bob’s speed when it passes through the lowest point of the swing? (c) What is the total energy stored in this oscillation assuming no losses?

(a) The frequency oscillates at \(f = 0.59\;{\rm{Hz}}\).

(b) The pendulum bob’s speed when it passes through the lowest point of the swing is \(0.56\;{\rm{m/s}}\).

(c) The total energy stored in this oscillation assuming no losses is \(4.5 \times {10^{ - 2}}\;{\rm{J}}\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Concept of frequency of SHM

In this problem, for calculating the frequency of the oscillation, apply the relation of the frequency with the gravitational acceleration and length of the pendulum.

Given data

The length of a simple pendulum is \(l = 0.72\;{\rm{m}}\).

The mass of the bob is \(m = 295\;{\rm{g}}\).

The initial angle of the bob with the length is \(\theta = 12^\circ \).

Calculation of frequency

The frequency of a simple pendulum is:

\(f = \frac{1}{{2\pi }}\sqrt {\frac{g}{l}} \)

Substitute the known values in the above relation to find frequency.

\(\begin{aligned}f &= \frac{1}{{2\pi }}\sqrt {\frac{{9.8\;{\rm{m/}}{{\rm{s}}^2}}}{{0.72\;{\rm{m}}}}} \\ &= 0.587\;{\rm{Hz}}\\\approx {\rm{0}}{\rm{.59}}\;{\rm{Hz}}\end{aligned}\)

Hence, the frequency of the simple pendulum is \(0.59\;{\rm{Hz}}\).

Calculation of velocity of bob

An illustrative figure is shown below:

The relation according to the conservation of energy is given by,

\(K{E_1} + P{E_1} = K{E_2} + P{E_2}\)

Here, kinetic energy of the bob initially is zero, \(K{E_1} = 0\), as well as the potential energy at the lowest point \(P{E_2} = 0\), since it will carry momentum due to moving downwards due to gravity.

On plugging the values in the above relation.

\(\begin{aligned}{c}0 + mg\left( {l - l\cos \theta } \right) &= \frac{1}{2}mv_2^2 + 0\\{v_2} &= \sqrt {2g\left( {l - l\cos \theta } \right)} \\{v_2} &= \sqrt {2\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {0.72\;{\rm{m}} - \left( {0.72\;{\rm{m}}} \right)\cos 12^\circ } \right)} \\{v_2} &= 0.56\;{\rm{m/s}}\end{aligned}\)

Hence, the velocity of the bob at its lowest point is \(0.56\;{\rm{m/s}}\).

Calculation of total energy

Total energy at its lowest point is equal to the kinetic energy at that point, since there will not be any potential energy left at the lowest point. Thus,

\(\begin{aligned}{c}{E_{\rm{T}}} = K{E_2} + P{E_2}\\ = \frac{1}{2}mv_2^2 + 0\\ = \frac{1}{2}mv_2^2\end{aligned}\)

Substitute the known values in the above relation to find total energy.

\(\begin{aligned}{c}{E_{\rm{T}}} &= \frac{1}{2}\left( {295\;{\rm{g}} \times \frac{{1\;{\rm{kg}}}}{{1000\;{\rm{g}}}}} \right){\left( {0.56\;{\rm{m/s}}} \right)^2}\\ &= 4.5 \times {10^{ - 2}}\;{\rm{J}}\end{aligned}\)

Hence, the total energy at its lowest point is \(4.5 \times {10^{ - 2}}\;{\rm{J}}\).

Most popular questions for Physics Textbooks

A \(0.650\;{\rm{kg}}\) mass oscillates according to the equation \(x = 0.25\sin \left( {4.70\;{\rm{t}}} \right)\) where \({\rm{x}}\) is in meters and \({\rm{t}}\) is in seconds. Determine (a) the amplitude, (b) the frequency, (c) the period, (d) the total energy, and (e) the kinetic energy and potential energy when \({\rm{x}}\) is 15 cm

Two balls are thrown off a building with the same speed, one straight up and one at a \({45^{\rm{o}}}\) angle. Which statement is true if air resistance can be ignored?

  1. Both hit the ground at the same time.
  2. Both hit the ground with the same speed.
  3. The one thrown at an angle hits the ground with a lower speed.
  4. The one thrown at an angle hits the ground with a higher speed.
  5. Both (a) and (b).

When the speed of your car is doubled, by what factor does its kinetic energy increase?

(a) \(\sqrt 2 \). (b) 2. (c) 4. (d) 8.

(II) A man doing push-ups pauses in the position shown in Fig. 9–65. His mass \({\bf{m = 68}}\;{\bf{kg}}\). Determine the normal force exerted by the floor (a) on each hand; (b) on each foot.

A diving board oscillates with simple harmonic motion of frequency \({\bf{2}}{\bf{.8}}\;{\bf{cycles per second}}\). What is the maximum amplitude with which the end of the board can oscillate in order that a pebble placed there (Fig. 11–56) does not lose contact with the board during the oscillation?
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