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Expert-verified Found in: Page 138 ### Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922 # 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 $$0.16\;{\rm{J}}$$.

See the step by step solution

## 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. ### Want to see more solutions like these? 