Simple Harmonic Motion

$\begin{aligned} & \frac{\mathrm{ML}^2}{3} \ddot{\theta}+\mathrm{K} \cdot \frac{\ell}{2} \theta \cdot \frac{\ell}{2}+\mathrm{K} \cdot \ell \cdot \theta \cdot \ell=0 \\ & \ddot{\theta}+\frac{15}{4} \frac{\mathrm{k}}{\mathrm{M}} \theta=0 \\ & \ddot{\theta}=-\left(\frac{15}{4} \frac{\mathrm{k}}{\mathrm{M}}\right) \theta \\ & \omega_1=\sqrt{\frac{15 \mathrm{~K}}{4 \mathrm{M}}}\end{aligned}$

$\frac{1}{3} \mathrm{ML}^2 \ddot{\theta}+2 \mathrm{~K} \frac{\mathrm{~L}}{2} \theta \cdot \frac{\mathrm{~L}}{2}=0$

$\begin{aligned} & \ddot{\theta}+\frac{3}{2} \frac{K}{M} \theta=0 \\ & \ddot{\theta}=-\frac{3}{2} \frac{K}{M} \theta \\ & \omega_2=\sqrt{\frac{3}{2} \frac{K}{M}} \\ & \frac{\omega_1}{\omega_2}=\sqrt{\frac{15}{4} \times \frac{2}{3}}=\sqrt{\frac{5}{2}}\end{aligned}$

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} $

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.

Here, ${x_1} = A\sin \omega t$

${x_2} = A\sin \left( {\omega t + {{2\pi } \over 3}} \right)$

$\therefore$ ${x_1} + {x_2} = A\sin \omega t + A\sin \left( {\omega t + {{2\pi } \over 3}} \right)$

$ = A\sin \omega t + A\left[ {\sin \omega t\cos {{2\pi } \over 3} + \cos \omega t\sin {{2\pi } \over 3}} \right]$

$ = A\sin \omega t + A\left[ {\sin \omega t\left( { - {1 \over 2}} \right) + \cos \omega t\left( {{{\sqrt 3 } \over 2}} \right)} \right]$

$ = {A \over 2}\sin \omega t + {{\sqrt 3 } \over 2}A\cos \omega t = A\left[ {\sin \omega t\cos {\pi \over 3} + \cos \omega t\sin {\pi \over 3}} \right]$

$ = A\sin \left( {\omega t + {\pi \over 3}} \right)$

$\because$ ${x_1} + {x_2} + {x_3} = 0$

$ \Rightarrow {x_3} = - ({x_1} + {x_2}) = - A\sin \left( {\omega t + {\pi \over 3}} \right)$

$ = A\sin \left( {\omega t + \pi + {\pi \over 3}} \right)$

${x_3} = A\sin \left( {\omega t + {{4\pi } \over 3}} \right)$

$\because$ ${x_3} = B\sin (\omega t + \phi )$

Hence, $B = A,\,\phi = {{4\pi } \over 3}$

The kinetic energy of the particle cannot be negative. The total energy E is the sum of the kinetic energy K(x) and potential energy V(x) i.e.,

E = K(x) + V(x). ....... (1)

From the given figure, V(x) $\ge$ 0 for all x. If E $\le$ 0 for some x then kinetic energy K(x) = E $-$ V(x) $\le$ 0 for those x and hence the motion of the particle is not allowed for those x. On the other hand, if E $\ge$ V₀ then K(x) = E $-$ V(x) $\ge$ 0 for all x and particle is allowed to move for all x, including infinity (in this case the particle will escape to infinity). Thus, for the particle to have a periodic motion, V₀ > E > 0. In this case, particle is allowed at points where

K(x) = E $-$ V(x) = E $-$ $\alpha$x⁴ $\ge$ 0 i.e.,

| x | $\le$ (E/$\alpha$)^1/4.

As $V = a{x^4}$

$[\alpha ] = {{[V]} \over {[{x^4}]}} = {{[M{L^2}{T^{ - 2}}]} \over {[{L^4}]}} = [M{L^{ - 2}}{T^{ - 2}}]$

By method of dimensions,

$\left[ {{1 \over A}\sqrt {{m \over \alpha }} } \right] = {{{{[M]}^{1/2}}} \over {[L]{{[M{L^{ - 2}}{T^{ - 2}}]}^{1/2}}}} = [T]$

Only option (b) has the dimensions of time.

For $|x| > {X_0}$, $V = {V_0}$ = constant

Force $ = - {{dV} \over {dx}} = 0$

Hence, acceleration of the particle is zero for $|x| > {X_0}$.

Since the restoring force is same in both springs (which are being in series), we have

${k_1}{x_1} = {k_2}{x_2}$

It is given that

${x_1} + {x_2} = A$

$ \Rightarrow {x_1} = {{A{k_2}} \over {{k_1} + {k_2}}}$

The restoring torque is

$J = - 2 \times kx\left( {{L \over 2}} \right)\cos \theta = I\left( {{{{d^2}\theta } \over {d{t^2}}}} \right)$

Now, $x = {L \over 2}\sin \theta $

Therefore, $J = - k\left( {{{{L^2}} \over 2}} \right)\sin \theta \cos \theta = I\left( {{{{d^2}\theta } \over {d{t^2}}}} \right)$

$ \Rightarrow \left( {{{ - k{L^2}} \over 4}} \right)\sin 2\theta = I\left( {{{{d^2}\theta } \over {d{t^2}}}} \right)$

For small $\theta$, $\sin 2\theta = 2\theta $. Therefore,

${{ - k{L^2}\theta } \over 2} = I\left( {{{{d^2}\theta } \over {d{t^2}}}} \right)$

where $I = {{M{L^2}} \over {12}}$. Therefore,

${{{d^2}\theta } \over {d{t^2}}} = \left( {{{ - 6k} \over M}} \right)\theta = - {\omega ^2}\theta $ (SHM)

$ \Rightarrow \omega = \sqrt {{{6k} \over M}} $

Hence, the frequency of oscillation is

${\omega \over {2\pi }} = {1 \over {2\pi }}\sqrt {{{6k} \over M}} $

The equation of single harmonic motion is $x = A\sin \left( {{{2\pi } \over T}t} \right)$, where A is amplitude and T is the time period of the motion.

From the given figure, $A = 1$ and $T = 8$. Thus,

$x = \sin \left( {{\pi \over 4}t} \right)$

The acceleration is

$a = {{{d^2}x} \over {d{t^2}}} = {{ - {\pi ^2}} \over {16}}\sin \left( {{\pi \over 4}t} \right)$

At time $t = {4 \over 3}s$, the acceleration is

$a = {{ - {\pi ^2}} \over {16}}\sin \left( {{\pi \over 4} \times {4 \over 3}} \right) = - {{\sqrt 3 } \over {32}}{\pi ^2}$ cm/s$^2$

Potential energy at point A. U$_\mathrm{A}$ = 0,

Kinetic Energy K$_\mathrm{A}$ = maximum

Potential energy at point B, U_B = maximum,

K_B = 0

The graph of potential energy as function of displacement of a simple pendulum will be parabolic graph as given in option p. Hence (A) $ \to $ (P). The choice (S) is also correct if the mean position of the pendulum is at the origin. Hence (A) $ \to $ (P), (S).

(P)

(S)

(B)

Slope of displacement time graph gives velocity.

$\therefore$ $v = {{ds} \over {dt}}$

If $\overrightarrow a = 0$ or constant

$v = u + at$

$v = 0 + at$

$\therefore$ $a = v/t$

$a = $ constant

(S)

$ \begin{aligned} & \text { (C) } \quad R=\frac{v^2 \sin 2 \theta}{g} \\ & \therefore \quad R \propto v^2 \rightarrow \operatorname{option}(\mathrm{S}) \end{aligned} $

(D) $T = 2\pi \sqrt {{l \over g}} $

$\therefore$ $T \propto \sqrt l $

${T^2} \propto l$ is a straight line

Given data:

• In the beginning, the time period of the simple pendulum is $ T_1 $.

• The point of suspension moves upward according to

  $ y = K t^2, \quad K = 1 \, \text{m/s}^2 $

• Thus, the acceleration of the support is:

  $ a = \frac{d^2y}{dt^2} = 2K = 2 \, \text{m/s}^2 $

• $ g = 10 \, \text{m/s}^2 $

We have to find the ratio:

$ \frac{T_1^2}{T_2^2} $

Step 1: Relation between acceleration and effective gravity

When the support is accelerated upward with acceleration $ a $, the effective gravity for the pendulum becomes:

$ g' = g + a $

Here,

$ a = 2 \, \text{m/s}^2 $

so,

$ g' = 10 + 2 = 12 \, \text{m/s}^2 $

Step 2: Relation of time period with gravity

For a simple pendulum:

$ T = 2\pi \sqrt{\frac{L}{g_{\text{eff}}}} $

Thus:

$ T_1 = 2\pi \sqrt{\frac{L}{g}} $

$ T_2 = 2\pi \sqrt{\frac{L}{g'}} $

Step 3: Ratio of squares of time periods

$ \frac{T_1^2}{T_2^2} = \frac{\frac{L}{g}}{\frac{L}{g'}} = \frac{g'}{g} $

Substitute values:

$ \frac{T_1^2}{T_2^2} = \frac{12}{10} = \frac{6}{5} $

✅ Final Answer:

$ \boxed{\frac{T_1^2}{T_2^2} = \frac{6}{5}} $

Option B

Here, A small body of mass $m$ attached to one end of a vertically hanging spring performs SHM.

Angular frequency $=\omega$, Amplitude $=a$.

Under SHM, velocity $v=\omega \sqrt{a^{2}-y^{2}}$ After detaching from spring, net downward acceleration of the block $=\mathrm{g}$

$\therefore$ Height attained by the block $=h$.

$\therefore \quad h=y+\frac{v^{2}}{2 g}$ or $h=y+\frac{\omega^{2}\left(a^{2}-y^{2}\right)}{2 g}$

For $h$ to be maximum, $\frac{d h}{d y}=0, y=y''$

$\therefore \quad \frac{d h}{d y}=1+\frac{\omega^{2}}{2 g}\left(-2 y^{\prime}\right)$

$0=\frac{1-2 \omega^{2} y^{\prime}}{2 g}$

$\frac{\omega^{2} y^{\prime}}{g}=1$

$y^{\prime}=\frac{g}{\omega^{2}}$

Since, $w^{2} > g$ (given)

$\therefore a > \frac{g}{\omega^{2}} \therefore a > y', y'$ from mean position $ < a$.

Hence, $y'=\frac{g}{\omega^{2}}$

The potential energy $U(x)=k\left[1-e^{-x^2}\right]$, varies with $x$ as shown in the figure.

The potential energy has a local minimum $\left(\frac{\mathrm{d} U}{\mathrm{~d} x}=\right.$ $0, \frac{\mathrm{~d}^2 U}{\mathrm{~d} x^2}>0$ ) at stable equilibrium and local maximum $\left(\frac{\mathrm{d} U}{\mathrm{~d} x}=0, \frac{\mathrm{~d}^2 U}{\mathrm{~d} x^2}<0\right)$ at unstable equilibrium. It is given that potential energy has a stable equilibrium at $x=0$. When particle is displaced from the origin, the force $F=-\mathrm{d} U / \mathrm{d} x$ brings it back. This force points towards the origin. There is no unstable equilibrium.

Total mechanical energy, $U+K=k / 2$, gives kinetic energy of the particle as $K=k / 2-U$, which is maximum at $x=0$. The force on a particle of mass $m$ placed near $x=0$ is given by

$ \begin{aligned} F & =-\frac{\mathrm{d} U}{\mathrm{~d} x}=-2 k x e^{-x^2}=-2 k x\left(1-\frac{x^2}{2!}+\cdots\right) \\\\ & \approx-2 k x \end{aligned} $

The force $F$ is proportional to $x$ and points towards $x=$ 0 . Thus, particle executes SHM with angular frequency $\omega=\sqrt{2 k / m}$.

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}}}$

The sharpness of resonance decreases with increasing column length.

The first resonance happens when length of air column $\approx$ $\lambda/2$. The prongs of the tuning fork are always kept in a vertical plane and amplitude of vibration is typically in the mm range.

From the figure, we have

${\lambda \over 4} = {l_1} + e$

where e is the end-correction. Therefore,

${l_1} + \left( {{\lambda \over 4} - e} \right) < {\lambda \over 4}$

$ \begin{aligned} & x=A \sin ^2 \omega t+B \cos ^2 \omega t+C \sin \omega t \cos \omega t \\ & x=\frac{A}{2}(1-\cos 2 \omega t)+\frac{B}{2}(1+\cos 2 \omega t)+\frac{C}{2} \sin 2 \omega t \\ & x=\frac{A+B}{2}+\left(\frac{B}{2}-\frac{A}{2}\right)+\cos 2 \omega t+\frac{C}{2} \sin 2 \omega t \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

$ \begin{aligned} & x=\frac{A+B}{2} \sqrt{\left(\frac{B-A}{2}\right)^2+\left(\frac{C}{2}\right)^2} \sin (2 \omega t+\alpha) \\ & \qquad \tan \alpha=\frac{B-A}{C} \\ & x-\frac{A+B}{2}=\sqrt{\left(\frac{B-A}{2}\right)^2+\left(\frac{C}{2}\right)^2} \sin (2 \omega t+\alpha) \\ & x-x_0=a_0 \sin (\omega t+\alpha) \\ & \text { for } A=-B, C=2 B . \\ & \text { Amplitude } a_0=\sqrt{\left(\frac{B-A}{2}\right)^2+\left(\frac{C}{2}\right)^2}=\sqrt{2} B \end{aligned} $

From (i)

(A) For $\mathrm{A}=\mathrm{B}=0, x=\frac{\mathrm{C}}{2} \quad \sin 2 \omega t(\mathrm{SHM})$

(B) For $\mathrm{A}=-\mathrm{B}, \mathrm{C}=2 \mathrm{~B}$

$ \begin{aligned} & x=\frac{-\mathrm{B}+\mathrm{B}}{2}+\left(+\frac{\mathrm{B}}{2}+\frac{\mathrm{B}}{2}\right) \cos 2 \omega t+\frac{2 \mathrm{~B}}{2} \sin 2 \omega t \\ & x=\cos 2 \omega t+\mathrm{B} \sin 2 \omega t \\ & x=\sqrt{2} \mathrm{~B} \sin \left(2 \omega t+\frac{\pi}{4}\right)(\mathrm{SHM}) \end{aligned} $

Resultant amplitude $=\sqrt{2}$

(C) For $\mathrm{A}=\mathrm{B}, \mathrm{C}=0$,

$ \begin{aligned} x & =\frac{A+A}{2}+\left(\frac{A}{2}-\frac{A}{2}\right) \cos 2 \omega t+0 \\ & =\frac{A+A}{2}=A(\text { no SHM }) \end{aligned} $

$ \begin{aligned} & \text { (D) } \mathrm{A}=\mathrm{B}, \mathrm{C}=2 \mathrm{~B} \\ & \begin{aligned} x=\frac{\mathrm{A}+\mathrm{A}}{2}+\left(\frac{\mathrm{A}}{2}-\frac{\mathrm{A}}{2}\right) \cos 2 \omega t+\frac{2 \mathrm{~B}}{2} \sin 2 \omega t & \\ =\mathrm{A}+\mathrm{B} \sin 2 \omega t(\mathrm{OR}) \\ x=\mathrm{B}+\mathrm{B} \sin 2 \omega t(\mathrm{SHM}), \text { Amplitude }=|\mathrm{B}| \end{aligned} \end{aligned} $