Simple Harmonic Motion

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

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} $

We know that, equation for SHM along x-axis is given by $x = A\sin (\omega t + \phi )$

Let mean position for 1st particle is at x = 0

So, the SHM equation for 1st particle,

${x_1} = A\sin (\omega t + {\phi _1})$

Now, as the separation between mean positions of both the particle is x$_0$

Let the ball of mass m be thrown up with an initial velocity u. Its velocity v and displacement x are related by v² $-$ u² = $-$2gx, where g is the acceleration due to gravity. The momentum (p = mv) is given by

p² = m²u² $-$ 2m²gx,

which gives

$p = \pm \sqrt {{m^2}{u^2} - 2{m^2}gx} $.

At x = 0, the momentum is mu when the ball starts going up and it becomes $-$mu when the ball comes back. At the maximum height, x = u²/(2g), the momentum becomes zero.

Energy of simple harmonic oscillator is

$E = {1 \over 2}k{A^2}$

where k is the force constant and A the amplitude of the oscillator. Since the oscillator is the same, the value of k is the same. Hence

${E_1} = {1 \over 2}kA_1^2$ and ${E_2} = {1 \over 2}kA_2^2$

$\therefore$ ${{{E_1}} \over {{E_2}}} = {\left( {{{{A_1}} \over {{A_2}}}} \right)^2}$

Now, A₁ = maximum value of displacement of oscillator having energy E₁ = 2a and A₂ = a. Therefore

${{{E_1}} \over {{E_2}}} = {\left( {{{2a} \over a}} \right)^2} = 4$. So, ${E_1} = 4{E_2}$

Due to upthrust, the spring will be compressed. Due to damping by the liquid, the final position will be smaller than the initial position. Hence choices (c) and (d) are not possible. Due to buoyancy, the block will move upwards. Hence, according to the given sign convention, position (x) is positive initially. When the system is released, x will decrease and momentum (p) will increase becoming maximum when the system reaches the mean position (x = 0) after which the momentum will decrease to zero when the oscillator reaches the extreme position, after which the momentum becomes negative. Hence the correct graph is (b).

The force exerted on charge +Q by the electric field $\overrightarrow E $ is

$\overrightarrow F = Q\overrightarrow E $

in the direction of $\overrightarrow E $. Since $\overrightarrow F $ is constant, a constant force is added to the applied force. Hence only the mean position will change.

The frequency will be same.

As ${v_0} = {1 \over {2\pi }}\sqrt {{k \over m}} $ does not depend on the constant external force.