Simple Harmonic Motion

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$^2$

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,

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

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

$ \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} $

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 time period of a simple harmonic motion (SHM) performed by a mass-spring system is given by the formula:

$T = 2\pi \sqrt{{M \over k}}$

where:

• T is the time period,
• M is the mass of the object, and
• k is the spring constant.

We know that if the mass is increased by m, the time period becomes $\frac{5T}{3}$. We can set up an equation for this new scenario:

$\frac{5T}{3} = 2\pi \sqrt{\frac{M + m}{k}}$

Since we know that $T = 2\pi \sqrt{\frac{M}{k}}$, we can substitute T in the equation above:

$\frac{5}{3} \cdot 2\pi \sqrt{\frac{M}{k}} = 2\pi \sqrt{\frac{M + m}{k}}$

Squaring both sides of the equation to eliminate the square root, we get:

$\frac{25}{9} \cdot \frac{M}{k} = \frac{M + m}{k}$

Solving for $\frac{m}{M}$, we get:

$\frac{m}{M} = \frac{25}{9} - 1 = \frac{16}{9}$

The period of a simple pendulum is given by:

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

where:

• T is the period,
• L is the length of the pendulum, and
• g is the acceleration due to gravity.

Since g is constant, we see that the period T is proportional to the square root of the length L.

If L is increased by 21%, the new length L' is L + 21%L = 1.21L. The new period T' is then:

$ T' = 2\pi \sqrt{\frac{L'}{g}} = 2\pi \sqrt{\frac{1.21L}{g}} = \sqrt{1.21}T \approx 1.1T $

The percentage increase in the time period is then:

$ \frac{T' - T}{T} \times 100\% = (\sqrt{1.21} - 1) \times 100\% \approx 10\% $

Therefore, the percentage increase in the time period of the pendulum of increased length is approximately 10%.

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 $\omega=\sqrt{2 k / m}$.