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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

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

(a) The amplitude of oscillation is \(0.25\;{\rm{m}}\).

(b) The frequency is \(0.75\;{\rm{Hz}}\).

(c) The period is \(1.34\;{\rm{s}}\).

(d) The total energy is \(0.45\;{\rm{J}}\).

(e) The kinetic energy and potential energy of the mass is \(0.29\;{\rm{J}}\) and \(0.16\;{\rm{J}}\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Concept of conservation of energy 

In this problem, the kinetic energy and potential energy are determined by applying the principle of conservation of total energy.

Given data

The displacement equation for SHM is \(x = 0.25\sin \left( {4.70t} \right)\).

The time is \(t\).

The mass is \({\rm{m}} = 0.650\;{\rm{kg}}\).

Calculation of amplitude

The amplitude of the oscillation will be equivalent to the maximum displacement. So, the amplitude is calculated as:

\(\begin{aligned}{c}A &= {x_{{\rm{max}}}}\\ &= 0.25\;{\rm{m}}\end{aligned}\)

Hence, amplitude of the mass is \(0.25\;{\rm{m}}\).

Calculation of frequency

The frequency of oscillation is calculated as:

\(f = \frac{\omega }{{2{\rm{\pi }}}}\)

Substitute the values in the above relation.

\(\begin{aligned}{c}f = \frac{{4.70\;{{\rm{s}}^{ - 1}}}}{{2{\rm{\pi }}}}\\f = 0.748\;{\rm{Hz}}\end{aligned}\)

Hence, the frequency of oscillation is 0.748 Hz

Calculation of period

The period is calculated as:

\(T = \frac{1}{f}\)

Substitute the values in the above relation.

\(\begin{aligned}{c}T = \frac{1}{{0.748\;{\rm{Hz}}}}\\T = 1.34\;{\rm{s}}\end{aligned}\)

Hence, the period of oscillation is 1.34 s.

Calculation of total energy

The total energy is calculated as:

\({{\rm{E}}_{\rm{T}}} = \frac{1}{2}{\rm{mv}}_{{\rm{max}}}^{\rm{2}}\)

Substitute the values in the above relation.

\begin{aligned}{c}{E_{\rm{T}}} &= \frac{1}{2}m{\left( {\omega A} \right)^2}\\{E_{\rm{T}}} &= \frac{1}{2}\left( {0.65\;{\rm{kg}}} \right){\left( {4.70\;{{\rm{s}}^{ - 1}}} \right)^2}{\left( {0.25\;{\rm{m}}} \right)^2}\\{E_{\rm{T}}} &= 0.45\;{\rm{J}}\end{aligned}

Hence, total energy of the mass is \(0.45\;{\rm{J}}\).

Calculation of kinetic and potential energy

According to the conservation of energy, total energy remains constant.

The potential energy is calculated as:

\begin{aligned}{c}PE &= \frac{1}{2}k{x^2}\\ &= \frac{1}{2}m{\left( {\omega x} \right)^2}\\ &= \frac{1}{2}\left( {0.65\;{\rm{kg}}} \right){\left( {4.70\;{{\rm{s}}^{ - 1}}} \right)^2}{\left( {0.15\;{\rm{m}}} \right)^2}\\ &= 0.16\;{\rm{J}}\end{aligned}

The kinetic energy is calculated as:

\begin{aligned}KE &= {E_{\rm{T}}} - PE\\ &= 0.45\;{\rm{J}} - {\rm{0}}{\rm{.16}}\;{\rm{J}}\\ &= 0.29\;{\rm{J}}\end{aligned}

Hence, potential and kinetic energy of the mass is \(0.16\;{\rm{J}}\) and \(0.29\;{\rm{J}}\), respectively.

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