Simple Harmonic Motion

Given, range of mass is from $m_1$ to $m_2$ i.e. 0 to $15 \mathrm{~kg}$

Length of spring, $x=0.25 \mathrm{~m}$

Time period of oscillation, $T=\frac{2 \pi}{5} \mathrm{~s}$

Let mass of body suspended $=m$

Spring constant $=k$

Acceleration due to gravity $g=10 \mathrm{~ms}^{-2}$

Since, $T =2 \pi \sqrt{\frac{m}{k}}$ ..... (i)

$\begin{aligned} \text{and} \quad F & =m g=k x \\ \Rightarrow \quad k & =\frac{m g}{x}=\frac{15 \times 10}{0.25} \mathrm{~N} / \mathrm{m} \end{aligned}$

Substituting in Eq. (i), we get

$\begin{aligned} & T=2 \pi \sqrt{\frac{m \times 25}{15 \times 10 \times 100}} \\ & \Rightarrow \quad\left(\frac{2 \pi}{5}\right)^2=(2 \pi)^2\left(\frac{m}{15 \times 40}\right) \\ & \Rightarrow \quad \frac{4 \pi^2}{25}=\frac{4 \pi^2 m}{15 \times 40} \Rightarrow m=24 \mathrm{~kg} \\ \end{aligned}$

Given, expansion of spring, $\Delta x=0.4 \mathrm{~m}$

Mass suspended in case $\mathbf{I}\left(m_1\right)$ and case II $\left(m_2\right)$ is $0.6 \mathrm{~kg}$ and $255 \mathrm{~g}$, respectively.

Acceleration due to gravity $g=10 \mathrm{~ms}^{-2}$

Since, force $F=m g=k \Delta x$

$\begin{aligned} & \therefore \text { Spring constant } k=\frac{m g}{\Delta x} \\ & \Rightarrow \quad k=\frac{m_1 g}{\Delta x_1}=\frac{0.6 \times 10}{0.4}=15 \mathrm{~N} / \mathrm{m} \end{aligned}$

and time period, $T=2 \pi \sqrt{\frac{m}{k}}$

$\begin{aligned} T_2 & =2 \pi \sqrt{\frac{m_2}{k}} \\ T_2 & =2 \pi \sqrt{\frac{0.255}{15}} \\ & =2 \pi \sqrt{\frac{255}{15 \times 10^3}} \\ & =2 \pi \sqrt{0.017}=0.82 \mathrm{~s} \end{aligned}$

Given, time period of vibration in air = T

Density of fluid = $\rho$/10

Time period of vibration in fluid = T′

Since, $T = 2\pi \sqrt {{m \over k}} $

and as the value of mass (m) and spring constant ( k) of spring-mass system is same.

$\therefore$ Time period will remain same.

Let k be the force constant of spring.

Time period, $T = 2\pi \sqrt {{M \over k}} $ ...... (i)

When block of mass m is placed in the tray, the time period of oscillations becomes

$T' = 2\pi \sqrt {{{M + m} \over k}} $ ...... (ii)

On dividing Eq. (ii) by Eq. (i), we get

$ \Rightarrow {{T'} \over T} = {{\sqrt {M + m} } \over {\sqrt M }}$

$ \Rightarrow {3 \over {1.5}} = \sqrt {{{12 + m} \over {12}}} $ [given, M = 12 kg, T = 1.5 s, T' = 3 s]

$ \Rightarrow \sqrt {{{12 + m} \over {12}}} = 2$

$\Rightarrow$ 12 + m = 48

$\Rightarrow$ m = 36 kg

Let displacement, x = A$\omega$ sin $\omega$t

Velocity, $v = {{dx} \over {dt}} = A\omega \cos \omega t$

Kinetic energy of particle in SHM,

$K = {1 \over 2}m{v^2} \Rightarrow K = {1 \over 2}m{\omega ^2}{A^2}{\cos ^2}\omega t$

$ = {1 \over 2}m{\omega ^2}{A^2}\left( {{{1 + \cos 2\omega t} \over 2}} \right)$ [$\because$ ${\cos ^2}\theta = {{1 + \cos 2\theta } \over 2}$]

$ = {1 \over 4}m{\omega ^2}{A^2}(1 + \cos 2\omega t)$

$\therefore$ New angular frequency, ${\omega _K} = 2\omega $

$\therefore$ Frequency of oscillation of kinetic energy = 2f [$\because$ $f = {\omega \over {2\pi }}$]

The frequency of oscillation

$ \begin{equation*} \omega=\sqrt{\frac{K}{m}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

The maximum acceleration in SHM,

$ \alpha_{\max }=\omega^2 \chi $

where, $x$ is the maximum extension in the spring.

To maintain contact with spring, the maximum SHM acceleration should be equal to acceleration due to gravity.

$ \text { } \quad \begin{align*} i.e., \omega^2 x & \left.=g \Rightarrow x=\frac{m g}{K} \text { [from Eq. (i)] }\right. \\ & =\frac{10 \times 10}{400}=\frac{1}{4}[\therefore K=400 \mathrm{~N} / \mathrm{m}] \\ & =0.25 \mathrm{~m} \text { or } 25 \mathrm{~cm} \end{align*} $

Given, amplitude,

$ A=0.2 \mathrm{~cm}=2 \times 10^{-3} \mathrm{~m} $

Velocity at mean position in SHM

$ v_{\max }=5 \mathrm{~m} / \mathrm{s} $

We know that,

$ \begin{aligned} v_{\max } & =A \omega \Rightarrow \omega=\frac{v_{\max }}{A} \\ & =\frac{5}{2 \times 10^{-3}}=25 \times 10^3 \\ & =2500 \mathrm{rad} \mathrm{~s}^{-1} \end{aligned} $

Displacement equation of simple harmonic motion is given as

$ \begin{aligned} x= & 6 \cos \left(2 \pi t+\frac{\pi}{3}\right) \\ \therefore \text { Velocity, } v & =\frac{d x}{d t}=\frac{d}{d t}\left[6 \cos \left(2 \pi t+\frac{\pi}{3}\right)\right] \\ & =-6 \sin \left(2 \pi t+\frac{\pi}{3}\right) \times 2 \pi \\ v & =-12 \pi \sin \left(2 \pi t+\frac{\pi}{3}\right) \end{aligned} $

$ \begin{aligned} & \therefore \text { Acceleration, } a=\frac{d v}{d t} \\ & =\frac{d}{d t}\left[-12 \pi \sin \left(2 \pi t+\frac{\pi}{3}\right)\right] \\ & =-12 \pi \cos \left(2 \pi t+\frac{\pi}{3}\right)(2 \pi) \end{aligned} $

$ \begin{aligned} & a=-24 \pi^2 \cos \left(2 \pi t+\frac{\pi}{3}\right) \\ & \text { At, } t=1 \mathrm{~s} \\ & a=-24 \pi^2 \cos \left(2 \pi+\frac{\pi}{3}\right) \\ & =-24 \pi^2 \cos \frac{\pi}{3}=-24 \pi^2 \times \frac{1}{2}=-12 \pi^2 \\ & \therefore|a|=12 \pi^2 \mathrm{~ms}^{-2} \end{aligned} $

Displacement of particle is

$ x=x_0 \cos \left(\omega t-\frac{\pi}{4}\right) $

Velocity, $v=\frac{d x}{d t}$

$ \Rightarrow v=-x_0 \omega \sin \left(\omega t-\frac{\pi}{4}\right) $

Acceleration, $a=\frac{d v}{d t}$

$ \begin{aligned} & =-x_0 \omega^2 \cos \left(\omega t-\frac{\pi}{4}\right) \\ & =x_0 \omega^2 \cdot \cos \left(\pi+\left(\omega t-\frac{\pi}{4}\right)\right) \\ & =x_0 \omega^2 \cdot \cos \left(\omega t+\frac{3 \pi}{4}\right) \\ & a=x_0 \omega^2 \cos \left(\omega t-\left(-\frac{3 \pi}{4}\right)\right) \end{aligned} $

Comparing with $a=A \cos (\omega t-\delta)$, we have

$ A=x_0 \omega^2, \delta=-\frac{3 \pi}{4} $

Average acceleration of particle in SHM between extreme and equilibrium position is

$ A_{\mathrm{avg}}=\frac{\omega^2 a}{\pi / 2}=\omega^2 a \times \frac{2}{\pi}\left(\mathrm{~ms}^{-2}\right) $

where, $\omega^2 a=A_{\text {max }}$

So, $\frac{A_{\text {avg }}}{A_{\text {max }}}=\frac{2}{\pi}$

Let a be the acceleration of the lift.

Given, the ratio of time periods of pendulum while the lift is moving upwards and downwards,

T₁ : T₂ = 1 : 3

When lift is moving upwards, then total time period,

${T_1} = 2\pi \sqrt {{I \over {g + a}}} $........ (i)

When lift is moving downwards, then total time period,

${T_2} = 2\pi \sqrt {{I \over {g - a}}} $ ....... (ii)

Dividing Eq. (i) by Eq. (ii), we get

${{{T_1}} \over {{T_2}}} = \sqrt {{{g - a} \over {g + a}}} \Rightarrow {1 \over 3} = \sqrt {{{g - a} \over {g + a}}} $

or, ${{g - a} \over {g + a}} = {\left( {{1 \over 3}} \right)^2}$ or, ${{g - a} \over {g + a}} = {1 \over 9}$

$ \Rightarrow 9g - 9a = g + a$

$ \Rightarrow 9g - g = 9a + a$

$ \Rightarrow 10a = 8g$

$a = {{8g} \over {10}} = {{8 \times 10} \over {10}} = 8$ m/s²

Potential of the given oscillator is

$V = {1 \over 2}k{(x - k)^2}$

Given: M = 10; m = 5, u = 1; k = 1

Initial momentum of the particle of mass m

= mu = m $\times$ 5 = 5m

Momentum of (oscillator + particle) after collision = (M + m)

Velocity of oscillator after collision = v

So, momentum of system = (M + m)v

From conservation of linear momentum, we have

(M + m) = mu = 5 $\times$ 1 = 5

For second collision, oscillator and particle have momentum in opposite direction.

Net or total momentum is zero.

Likewise after 4^th, 6^th, 8^th, 10^th, 12^th collision the momentum is zero. After 12^th collision, Mass of oscillator and 12 particles will be (10 + 12 $\times$ 5) = 70

Now, from conservation of linear momentum, for 13^th collision, we have

$70 \times 0 + 5 \times 1 = (70 + 5)v' \Rightarrow v' = {5 \over {75}} \Rightarrow {1 \over {15}}$

Total mass after 13^th collision = (10 + 13 $\times$ 5) = 75

Kinetic energy of system $ = {1 \over 2}mv{'^2}$

$ \Rightarrow KE = {1 \over 2} \times 75 \times {1 \over {15}} \times {1 \over {15}}$

$ \Rightarrow {1 \over 2}k{A^2} = {1 \over 2} \times {{75} \over {225}} = {1 \over 6}$

$ \Rightarrow {1 \over 2} \times 1 \times {A^2} = {1 \over 6} \Rightarrow {A^2} = {1 \over 3}$

$ \Rightarrow A = {1 \over {\sqrt 3 }}$

The given simple harmonic motions to form Lissajous figures are $x(t) = A\sin (at + \delta )$ and $y(t) = B\sin (bt)$.

For parabola, conditions should be

A = B or A $\ne$ B, a = 2b, $\delta$ = $\pi$/2

For line, conditions should be

A = B, a = b, $\delta$ = $\pi$

For circle, condition should be

A = B, a = b; $\delta$ = $\pi$/2

For ellipse, condition should be

A $\ne$ B, a = b; $\delta$ = $\pi$/2

Therefore, we obtain an ellipse.

To determine the time taken for the mechanical energy of the damped oscillator to drop to half its initial value, we'll use the principles of damped harmonic motion.

Given:

Mass of the block ($ m $): 0.1 kg

Spring constant ($ k $): 640 N/m

Damping constant ($ b $): $ 10^{-2} $ kg/s

Understanding Damped Harmonic Motion:

In damped harmonic motion, the amplitude of oscillation decreases exponentially over time due to the damping force. The mechanical energy ($ E $) of the oscillator is proportional to the square of its amplitude ($ A $):

$ E(t) \propto A(t)^2 $

The amplitude as a function of time is given by:

$ A(t) = A_0 \, e^{- \frac{b}{2m} t} $

Where:

$ A_0 $ is the initial amplitude.

$ b $ is the damping constant.

$ m $ is the mass.

$ t $ is the time.

Therefore, the mechanical energy as a function of time is:

$ E(t) = E_0 \, e^{- \frac{b}{m} t} $

Where $ E_0 $ is the initial mechanical energy.

Calculating the Time When Energy Drops to Half:

We need to find the time $ t $ when $ E(t) = \frac{1}{2} E_0 $:

$ \frac{1}{2} E_0 = E_0 \, e^{- \frac{b}{m} t} $

Simplify:

$ \frac{1}{2} = e^{- \frac{b}{m} t} $

Take the natural logarithm of both sides:

$ \ln\left(\frac{1}{2}\right) = - \frac{b}{m} t $

Simplify $ \ln\left(\frac{1}{2}\right) = -\ln(2) $:

$ - \ln(2) = - \frac{b}{m} t $

Cancel negatives:

$ \ln(2) = \frac{b}{m} t $

Solve for $ t $:

$ t = \frac{m}{b} \ln(2) $

Plugging in the Given Values:

$ t = \frac{0.1\, \text{kg}}{10^{-2}\, \text{kg/s}} \ln(2) $

Calculate $ \ln(2) $:

$ \ln(2) \approx 0.6931 $

Now compute $ t $:

$ t = \frac{0.1}{0.01} \times 0.6931 = 10 \times 0.6931 = 6.931\, \text{s} $

Conclusion:

The time taken for the mechanical energy to drop to half its initial value is approximately 6.93 seconds, which is closest to 7 seconds among the given options.

1st Particle

P = 0 at x = a $\Rightarrow$ 'a' is the amplitude of oscillation 'A₁'.

At x = 0, P = b (at mean position)

$ \Rightarrow m{v_{\max }} = b \Rightarrow {v_{\max }} = {b \over m}$

${E_1} = {1 \over 2}mv_{\max }^2 = {m \over 2}{\left[ {{b \over m}} \right]^2} = {{{b^2}} \over {2m}}$

${A_1}{\omega _1} = {v_{\max }} = {b \over m}$

$ \Rightarrow {\omega _1} = {b \over {ma}} = {1 \over {m{n^2}}}(A = a,\,{a \over b} = {n^2})$

2nd Particle

P = 0 at x = R $\Rightarrow$ A₂ = R

At x = 0, P = R $\Rightarrow$ ${v_{\max }} = {R \over m}$

${E_2} = {1 \over 2}mv_{\max }^2 = {m \over 2}{\left[ {{R \over m}} \right]^2} = {{{R^2}} \over {2m}}$

${A_2}{\omega _2} = {R \over m} \Rightarrow {\omega _2} = {R \over {mR}} = {1 \over m}$

(b) ${{{\omega _2}} \over {{\omega _1}}} = {{1/m} \over {1/m{n^2}}} = {n^2}$

(c) ${\omega _1}{\omega _2} = {1 \over {m{n^2}}} \times {1 \over m} = {1 \over {{m^2}{n^2}}}$

(d) ${{{E_1}} \over {{\omega _1}}} = {{{b^2}/2m} \over {1/m{n^2}}} = {{{b^2}{n^2}} \over 2} = {{{a^2}} \over {2{n^2}}} = {{{R^2}} \over 2}$

${{{E_2}} \over {{\omega _2}}} = {{{R^2}/2m} \over {1/m}} = {{{R^2}} \over 2}$

$ \Rightarrow {{{E_1}} \over {{\omega _1}}} = {{{E_2}} \over {{\omega _2}}}$

Since, the block starts executing simple harmonic motion from extreme position, we have

$x = a\cos \omega t$

At 1 s, we have

$x = 0.2\cos \left( {{\pi \over 3}} \right) = 0.1\,m$

That is, the block is at a distance 5.1 $-$ 0.10 = 5 m from O, which is the range of the bebble as well. Now,

$R = {{{v^2}\sin 2\alpha } \over g}$

$S = {{{v^2}\sin 2(\pi /4)} \over {10}}$

Therefore, $v = \sqrt {50} $ m/s

Alternate Method :

Since, pebble strikes the oscillating mass after 1 sec., its time of flight is $1 \mathrm{~s}$

$ \begin{aligned} & 1=\frac{2 v \sin 45^{\circ}}{\mathrm{g}}=\frac{\sqrt{2} v}{10} \\\\ & v=\frac{10}{\sqrt{2}}=5 \sqrt{2}=\sqrt{50} \mathrm{~m} / \mathrm{s} \end{aligned} $