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Found in: Page 441

Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

A particle executes linear SHM with frequency ${\mathbf{0}}{\mathbf{.}}{\mathbf{25}}{\mathbf{}}{\mathbf{Hz}}$ about the point ${\mathbf{x}}{\mathbf{=}}{\mathbf{0}}$. At ${\mathbf{t}}{\mathbf{=}}{\mathbf{0}}$, it has displacement ${\mathbf{x}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{37}}{\mathbit{c}}{\mathbit{m}}$ and zero velocity. For the motion, determine the (a) period, (b) angular frequency, (c) amplitude, (d) displacement x(t), (e) velocity v(t), (f) maximum speed, (g) magnitude of the maximum acceleration, (h) displacement at ${\mathbf{t}}{\mathbf{=}}{\mathbf{3}}{\mathbf{.}}{\mathbf{0}}{\mathbf{s}}$, and (i) speed at ${\mathbf{t}}{\mathbf{=}}{\mathbf{30}}{\mathbf{s}}$.

1. The period (T) is $4\mathrm{sec}$.
2. The angular frequency ($\omega$) is $1.57\mathrm{rad}/\mathrm{sec}$.
3. The amplitude (${x}_{m}$) is $0.37\mathrm{cm}$.
4. The displacement is $x\left(t\right)=\left(0.37cm\right)\mathrm{cos}\left(05\left(1.57t\right)$.
5. The velocity is $v\left(t\right)=-\left(0.58\mathrm{cm}/\mathrm{s}\right)\mathrm{sin}\left(1.57t\right)$.
6. The maximum speed is $0.58\mathrm{cm}/\mathrm{s}$.
7. The magnitude of maximum acceleration is $0.91\mathrm{cm}/{\mathrm{s}}^{2}$.
8. The displacement at $t=3$ is 0.

9.The speed at $t=3\mathrm{s}$ is $0.58\mathrm{cm}/\mathrm{s}$.

See the step by step solution

Step 1: The given data

• The frequency of oscillation is, $\mathrm{f}=0.25\mathrm{Hz}$.
• At the time $t=0$, velocity is, $v=0\mathrm{m}/\mathrm{s}$.
• At $\mathrm{t}=0$, the displacement is, $x=0.37\mathrm{cm}\mathrm{or}0.0037\mathrm{m}$.

Step 2: Understanding the concept of SHM

As per Hooke’s law, a particle with mass m moves under the influence of restoring force, ${\mathbit{F}}{\mathbf{=}}{\mathbf{-}}{\mathbit{k}}{\mathbit{x}}$ undergoes a simple harmonic motion. Here, F is restoring force, k is force constant and x is the displacement from the mean position.

In simple harmonic motion, displacement of the particle is given by the equation,

${\mathbit{x}}{\mathbf{=}}{{\mathbit{x}}}_{{\mathbf{m}}}{\mathbf{cos}}{\mathbf{\left(}}{\mathbit{\omega }}{\mathbit{t}}{\mathbf{+}}{\mathbf{\phi }}{\mathbf{\right)}}$

Using the expression for a simple harmonic motion for displacement, velocity, and formulae for ${\mathbit{T}}{\mathbf{,}}{\mathbit{\omega }}{\mathbf{,}}{{\mathbit{V}}}_{{\mathbf{max}}}{\mathbf{,}}{{\mathbit{a}}}_{{\mathbf{max}}}$ we can find the respective values.

Formula:

The angular frequency of oscillation, $\omega =2\mathrm{\pi }f$ (i)

The period of oscillation, $T=\frac{1}{f}$ (ii)

The maximum speed of a body, ${v}_{max}=\omega {x}_{m}$ (iii)

The magnitude of maximum acceleration of a body, $\left|{a}_{max}\right|={\omega }^{2}{x}_{m}$ (iv)

The displacement equation at zero phase, $x\left(t\right)={x}_{m}\mathrm{cos}\left(\omega t\right)$ (v)

The velocity equation at zero phase, $v\left(t\right)=-\omega {x}_{m}\mathrm{sin}\left(\omega t\right)$ (vi)

Here, $f$ is frequency, ${x}_{m}$ is maximum displacement or amplitude, $t$ is time.

Step 3: (a) Calculation of period

Using equation (ii), we can get the period of oscillations as:

$T=\frac{1}{0.25\mathrm{Hz}}\phantom{\rule{0ex}{0ex}}=4\mathrm{s}$

Hence, the value of period is $4\mathrm{s}$.

Step 4: (b) Calculation of angular frequency

Using equation (i), we can get the angular frequency of oscillation as:

$\omega =2×3.14×0.25\mathrm{Hz}\phantom{\rule{0ex}{0ex}}=1.57\mathrm{rad}/\mathrm{sec}$

Hence, the value of angular frequency is role="math" localid="1657257255516" $1.57\mathrm{rad}/\mathrm{sec}$.

Step 5: (c) Calculation of amplitude

As at $x=0.0037\mathrm{m},v=0$ and at extreme positions $v=0$

role="math" localid="1657257658285" $\mathrm{Hence},\mathrm{x}={\mathrm{x}}_{m}\phantom{\rule{0ex}{0ex}}=3.7×{10}^{-3}\mathrm{m}\phantom{\rule{0ex}{0ex}}=0.0037\mathrm{m}$

Hence, the value of amplitude is $0.0037\mathrm{m}$.

Step 6: (d) Calculation of the displacement equation

By substituting the given and derived values in equation (v), we get the displacement equation as:

$x\left(t\right)=\left(0.37\mathrm{cm}\right)\mathrm{cos}\left(1.57\mathrm{t}\right)$

Hence, the displacement equation is $\left(0.37\mathrm{cm}\right)\mathrm{cos}\left(1.57\mathrm{t}\right)$.

Step 7: (e) Calculation of velocity equation

By substituting the given and derived values in equation (vi), we get the velocity equation as:

$v\left(t\right)=-1.57\mathrm{rad}/\mathrm{s}×0.37\mathrm{cm}\mathrm{sin}\left(1.57t\right)\phantom{\rule{0ex}{0ex}}=-\left(0.58\mathrm{cm}/\mathrm{s}\right)\mathrm{sin}\left(1.57t\right)$

Hence, the velocity equation is $-\left(0.58cm/s\right)\mathrm{sin}\left(1.57t\right)$.

Step 8: (f) Calculation of maximum speed

Using equation (iii), we can get the maximum speed of the particle as:

${\mathrm{v}}_{max}=1.57\mathrm{rad}/\mathrm{s}×0.37\mathrm{cm}\phantom{\rule{0ex}{0ex}}=0.58\mathrm{cm}/\mathrm{s}$

Hence, the value of maximum speed is $0.58\mathrm{cm}/\mathrm{s}$.

Step 9: (g) Calculation of maximum acceleration

Using equation (iv), we can get the magnitude of the maximum acceleration as:

$\left|{\mathrm{a}}_{\mathrm{max}}\right|={\left(1.57\mathrm{rad}/\mathrm{s}\right)}^{2}×0.37\mathrm{cm}\phantom{\rule{0ex}{0ex}}=0.91\mathrm{cm}/{\mathrm{s}}^{2}$

Hence, the value of maximum acceleration is $0.91\mathrm{cm}/{\mathrm{s}}^{2}$.

Step 10: (h) Calculation of displacement at  t=3 s

Substituting the given values for in equation (v), we get the displacement as:

$\mathrm{x}\left(3\right)=\left(0.37\mathrm{cm}\right)\mathrm{cos}\left(1.57\mathrm{rad}/\mathrm{s}×3\mathrm{s}\right)\phantom{\rule{0ex}{0ex}}=\left(0.37\mathrm{cm}\right)\mathrm{cos}\left(4.71\mathrm{rad}\right)\phantom{\rule{0ex}{0ex}}\approx 0$

Hence, the displacement value is 0 m.

Step 11: (i) Calculation of velocity at t = 3 s

Substituting the given values for in equation (vi), we get the velocity as:

$\mathrm{v}\left(3s\right)=-\left(1.57\mathrm{rad}/\mathrm{s}\right)×\left(0.37\mathrm{cm}\right)sin\left(1.57\mathrm{rad}/\mathrm{s}×3\mathrm{s}\right)\phantom{\rule{0ex}{0ex}}=\left(-0.5809\mathrm{cm}/\mathrm{s}\right)×sin\left(4.71\mathrm{rad}\right)\phantom{\rule{0ex}{0ex}}=\left(-0.5809\mathrm{cm}/\mathrm{s}\right)×\left(-0.999\right)\phantom{\rule{0ex}{0ex}}=0.58\mathrm{cm}/\mathrm{s}$

Hence, the displacement value is 0.58 cm/s.