Chapter 13: Oscillations

Ans: In the above question it is given that:

\[Mass\text{ }of\text{ }the\text{ }automobile,\text{ }m\text{ }=\text{ }3000\text{ }kg\]

\[Displacement\text{ }in\text{ }the\text{ }suspension\text{ }system,\text{ }x\text{ }=\text{ }15\text{ }cm\text{ }=\text{ }0.15\text{ }m\]

There are 4 springs in parallel to the support of the mass of the automobile.

The equation for the restoring force for the system is \[F\text{ }=\text{ }4kx\text{ }=\text{ }mg\].

Where, k is the spring constant of the suspension system.

Time period, \[T=2\pi \sqrt{\frac{m}{4k}}\]

And

$k=mg/4x=3000\times 10/4\times 0.15=5000=5\times {{10}^{4}}Nm$

Spring Constant, $k=5\times {{10}^{4}}Nm$.

1. the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.

Ans: We know that each wheel supports a mass, \[M\text{ }=\text{ }3000/4\text{ }=\text{ }750\text{ }kg\].

For damping factor b, the equation for displacement is written as:

$x={{x}_{0}}{{e}^{-bt/2M}}$

The amplitude of oscillation decreases by \[50\text{ }%\].

$x={{x}_{0}}/2$

$\Rightarrow {{x}_{0}}/2={{x}_{0}}{{e}^{-bt/2M}}$

${{\log }_{e}}2=bt/2M$

$b=2M{{\log }_{e}}2/t$

Where, 

\[T=2\pi \sqrt{\frac{m}{4k}}=2\pi \sqrt{\frac{3000}{4\times 5\times {{10}^{4}}}}=0.7691s\]

$\Rightarrow b=\frac{2\times 750\times 0.693}{0.7691}=1351.53kg/s$

Hence, the damping constant of the spring is $1351.53kg/s$.

Ans: Consider,

The equation of displacement of a particle executing SHM at an instant t is given as:

$x=A\sin \omega t$

where,

A is the amplitude of oscillation,

\[\text{Angular frequency=}\omega \text{=}\sqrt{\frac{k}{M}}\]

The velocity of the particle is given by $v=dx/dt=A\omega \cos \omega t$.

The kinetic energy of the particle is:

${{E}_{k}}=1/2M{{v}^{2}}=1/2M{{A}^{2}}{{\omega }^{2}}{{\cos }^{2}}\omega t$

The potential energy of the particle is:

${{E}_{p}}=1/2k{{x}^{2}}=1/2{{M}^{2}}{{\omega }^{2}}{{A}^{2}}{{\sin }^{2}}\omega t$

For time period T, the average kinetic energy over a single cycle is given as:

${{\left( {{E}_{k}} \right)}_{Avg}}=\frac{1}{T}\int\limits_{0}^{T}{{{E}_{k}}dt}$

\[=\frac{1}{T}\int\limits_{0}^{T}{\frac{1}{2}M{{A}^{2}}{{\omega }^{2}}co{{s}^{2}}\omega tdt}\]

\[=\frac{1}{2T}M{{A}^{2}}{{\omega }^{2}}\left[ 1+\frac{\sin 2\omega t}{2\omega } \right]_{0}^{T}\]

\[=\frac{1}{2T}M{{A}^{2}}{{\omega }^{2}}\left( T \right)\]

\[=\frac{1}{4}M{{A}^{2}}{{\omega }^{2}}\] ……. (i)

And, average potential energy over one cycle is given as:

${{\left( {{E}_{p}} \right)}_{Avg}}=\frac{1}{T}\int\limits_{0}^{T}{{{E}_{p}}dt}$

$=\frac{1}{T}\int\limits_{0}^{T}{\frac{1}{2}M{{A}^{2}}{{\omega }^{2}}{{\sin }^{2}}\omega tdt}$

$=\frac{1}{2T}M{{\omega }^{2}}{{A}^{2}}\int\limits_{0}^{T}{\frac{\left( 1+\cos 2\omega t \right)}{2}}dt$

$=\frac{1}{4T}M{{\omega }^{2}}{{A}^{2}}\left( T \right)$

$=\frac{M{{\omega }^{2}}{{A}^{2}}}{4}$ …… (ii)

It can be interpreted from equations (i) and (ii) that the average kinetic energy for a given time period is equal to the average potential energy for the same time period.

Ans: In the above question it given that:

\[Mass\text{ }of\text{ }the\text{ }circular\text{ }disc,\text{ }m\text{ }=\text{ }10\text{ }kg\]

\[Radius\text{ }of\text{ }the\text{ }disc,\text{ }r\text{ }=\text{ }15\text{ }cm\text{ }=\text{ }0.15\text{ }m\]

\[The\text{ }torsional\text{ }oscillations\text{ }of\text{ }the\text{ }disc\text{ }has\text{ }a\text{ }time\text{ }period,\text{ }T\text{ }=\text{ }1.5\text{ }s\]

The moment of inertia of the disc is:

$I=1/2m{{r}^{2}}$

$=1/2\left( 10 \right){{\left( 0.15 \right)}^{2}}$

$=0.1125kg/{{m}^{2}}$

Time period,

$T=2\pi \sqrt{\frac{I}{\alpha }}$

And

 $\alpha =4{{\pi }^{2}}I/{{T}^{2}}$ 

$=4{{\pi }^{2}}\left( \frac{0.1125}{{{1.5}^{2}}} \right)$

$=1.972N/rad$

Therefore, the torsional spring constant of the wire is $1.972N/rad$.

Ans: In the above question it is given that:

Amplitude, r= 5 cm = 0.05 m

$r=5cm=0.05m$

Time period, \[T\text{ }=\text{ }0.2\text{ }s\].

$\omega =2\pi /T=2\pi /0.2=1\text{0 }\!\!\pi\!\!\text{  rad/s}$

When displacement is y, then acceleration, $A=-{{\omega }^{2}}y$.

Velocity is $V=\omega \sqrt{{{r}^{2}}-{{y}^{2}}}$

Now,

Case (a) When \[y\text{ }=\text{ }5cm\text{ }=\text{ }0.05\text{ }m\]

$A=-{{\left( 10\pi  \right)}^{2}}\times 0.05=-5{{\pi }^{2}}m/{{s}^{2}}$

$V=10\pi \sqrt{{{\left( 0.05 \right)}^{2}}-{{\left( 0.05 \right)}^{2}}}=0$

Case (b) When \[y\text{ }=\text{ 3}cm\text{ }=\text{ }0.03\text{ }m\]

$A=-{{\left( 10\pi  \right)}^{2}}\times 0.03=-3{{\pi }^{2}}m/{{s}^{2}}$

$V=10\pi \sqrt{{{\left( 0.05 \right)}^{2}}-{{\left( 0.03 \right)}^{2}}}=0.4\pi m/s$

Case (c) When \[y\text{ }=\text{ 0}cm\text{ }=\text{ }0\text{ }m\]

$A=0m/{{s}^{2}}$

$V=10\pi \sqrt{{{\left( 0.05 \right)}^{2}}-{{\left( 0 \right)}^{2}}}=0.5\pi m/s$

Ans: In the above question it is given that:

$x=a\cos \left( \omega t+\theta  \right)$

Where,

A is the amplitude,

x is the displacement,

$\theta $ is the phase constant.

Then.

\[v=dx/dt=-A\omega \sin \left( \omega t+\theta  \right)\] 

At $t=0$, $x={{x}_{0}}$

${{x}_{0}}=A\cos \theta $         …… (i)

And $dx/dt=-{{v}_{0}}=A\omega \sin \theta $

$A\sin \theta ={{v}_{0}}/\omega $ …… (ii)

Squaring and adding (i) an (ii),

${{A}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta  \right)={{x}_{0}}^{2}+\left( \frac{{{v}_{0}}^{2}}{{{\omega }^{2}}} \right)$

$\Rightarrow A=\sqrt{{{x}_{0}}^{2}+{{\left( \frac{{{v}_{0}}}{\omega } \right)}^{2}}}$

Thus, the resulting amplitude is $\sqrt{{{x}_{0}}^{2}+{{\left( \frac{{{v}_{0}}}{\omega } \right)}^{2}}}$