Oscillations

A motion is periodic if it repeats itself at regular intervals of time.

(a) Swimmer going from one bank to the other and back:
The swimmer's trip is a to-and-fro motion, but the time taken for each trip need not be the same (depends on the swimmer's effort, current, etc.). This motion does not repeat at regular time intervals. Hence it is not periodic.

(b) Freely suspended bar magnet displaced from N-S direction and released:
The magnet experiences a restoring torque due to Earth's magnetic field and oscillates about the N-S direction. It repeats its motion at regular intervals. Hence it is periodic (and approximately SHM for small displacements).

(c) Hydrogen molecule rotating about its centre of mass:
The rotation repeats itself after every complete revolution at regular time intervals. Hence it is periodic.

(d) Arrow released from a bow:
After release, the arrow moves in a parabolic trajectory under gravity and does not repeat its motion. Hence it is not periodic.

For SHM, the restoring force must be proportional to displacement and directed towards the equilibrium position: $F = -kx$.

(a) Rotation of earth about its axis:
This is a uniform circular motion (periodic) but the restoring force condition $F = -kx$ is not satisfied. It is periodic but not SHM.

(b) Oscillating mercury column in a U-tube:
When mercury is displaced, the restoring force is proportional to the displacement of the mercury column. This satisfies $F = -kx$. Hence it is (nearly) SHM.

(c) Ball bearing inside a smooth curved bowl (released slightly above lowest point):
For small displacements from the lowest point, the restoring force is proportional to displacement (similar to a simple pendulum). Hence it is (nearly) SHM.

(d) General vibrations of a polyatomic molecule:
A polyatomic molecule has many degrees of freedom and its vibrations are a superposition of many modes. The restoring force is not simply proportional to a single displacement. Hence it is periodic but not SHM (it is a superposition of multiple SHMs).

Note: The figures are not fully visible in the OCR, but based on the standard NCERT problem, the four plots are described as follows and the standard answers are given.

(a) Plot (a): The graph shows a sinusoidal (or repeating) pattern. This represents periodic motion. The motion repeats every 2 s, so $T = 2\,\text{s}$.

(b) Plot (b): The graph shows a non-repeating pattern (the displacement keeps increasing or is irregular). This is not periodic.

(c) Plot (c): The graph shows a repeating pattern. This represents periodic motion with period $T = 4\,\text{s}$.

(d) Plot (d): The graph shows a repeating pattern. This represents periodic motion with period $T = 2\,\text{s}$.

Summary:
- Plot (a): Periodic, $T = 2\,\text{s}$
- Plot (b): Not periodic
- Plot (c): Periodic, $T = 4\,\text{s}$
- Plot (d): Periodic, $T = 2\,\text{s}$

Concept: A function represents SHM if it can be written in the form $A\sin(\omega t + \phi)$ or $A\cos(\omega t + \phi)$. It is periodic but not SHM if it repeats but cannot be written in that single sinusoidal form. It is non-periodic if it does not repeat.

---

(a) $\sin\omega t - \cos\omega t$

Write in standard form:
$$\sin\omega t - \cos\omega t = \sqrt{2}\left(\frac{1}{\sqrt{2}}\sin\omega t - \frac{1}{\sqrt{2}}\cos\omega t\right) = \sqrt{2}\sin\left(\omega t - \frac{\pi}{4}\right)$$

This is of the form $A\sin(\omega t + \phi)$, so it represents SHM.

$$\boxed{T = \frac{2\pi}{\omega}}$$

---

(b) $\sin^3\omega t$

Using the identity: $\sin^3\theta = \frac{3\sin\theta - \sin 3\theta}{4}$

$$\sin^3\omega t = \frac{3\sin\omega t - \sin 3\omega t}{4}$$

This is a superposition of two sinusoids with frequencies $\omega$ and $3\omega$. It is periodic but not SHM.

The period is the LCM of $\dfrac{2\pi}{\omega}$ and $\dfrac{2\pi}{3\omega}$, which is $\dfrac{2\pi}{\omega}$.

$$\boxed{T = \frac{2\pi}{\omega}}$$

---

(c) $3\cos\left(\dfrac{\pi}{4} - 2\omega t\right)$

Using $\cos(-\theta) = \cos\theta$:
$$3\cos\left(\frac{\pi}{4} - 2\omega t\right) = 3\cos\left(2\omega t - \frac{\pi}{4}\right)$$

This is of the form $A\cos(\omega' t + \phi)$ with $\omega' = 2\omega$. It represents SHM.

$$\boxed{T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}}$$

---

(d) $\cos\omega t + \cos 3\omega t + \cos 5\omega t$

This is a superposition of three sinusoids with frequencies $\omega$, $3\omega$, and $5\omega$. Each is periodic. The overall period is the LCM of $\dfrac{2\pi}{\omega}$, $\dfrac{2\pi}{3\omega}$, $\dfrac{2\pi}{5\omega}$, which is $\dfrac{2\pi}{\omega}$.

It is periodic but not SHM.

$$\boxed{T = \frac{2\pi}{\omega}}$$

---

(e) $\exp(-\omega^2 t^2)$

This is a Gaussian function. As $t \to \infty$, the function $\to 0$. It never repeats. It is non-periodic (no period).

---

(f) $1 + \omega t + \omega^2 t^2$

This is a polynomial in $t$. As $t$ increases, the function increases without bound and never repeats. It is non-periodic (no period).

Setup: The particle oscillates between A and B, 10 cm apart. The midpoint O is the equilibrium (mean) position. Taking A→B as positive direction:
- A is at $x = -5\,\text{cm}$ (left extreme)
- O is at $x = 0$ (mean position)
- B is at $x = +5\,\text{cm}$ (right extreme)

Key rules for SHM:
- At extreme positions (A or B): velocity = 0
- Acceleration and force are always directed towards the mean position O (i.e., $a = -\omega^2 x$)
- Velocity is positive when moving from A to B (positive direction), negative when moving from B to A

---

(a) At end A ($x = -5\,\text{cm}$):
- Velocity: zero (extreme position), $v = 0$
- Acceleration: directed from A towards O, i.e., in positive direction → positive
- Force: same direction as acceleration → positive

$$v = 0,\quad a = +,\quad F = +$$

---

(b) At end B ($x = +5\,\text{cm}$):
- Velocity: zero (extreme position), $v = 0$
- Acceleration: directed from B towards O, i.e., in negative direction → negative
- Force: same direction as acceleration → negative

$$v = 0,\quad a = -,\quad F = -$$

---

(c) At mid-point O going towards A:
- Position: $x = 0$ (mean position)
- Moving towards A means moving in negative direction → velocity is negative
- At mean position, $a = -\omega^2 \times 0 = 0$ → acceleration is zero
- Force = 0

$$v = -,\quad a = 0,\quad F = 0$$

---

(d) At 2 cm away from B going towards A:
- B is at $x = +5\,\text{cm}$; 2 cm away from B towards A means $x = +5 - 2 = +3\,\text{cm}$
- Moving towards A → velocity is negative
- $a = -\omega^2(+3)$ → acceleration is negative (towards O, i.e., towards A)
- Force is negative

$$v = -,\quad a = -,\quad F = -$$

---

(e) At 3 cm away from A going towards B:
- A is at $x = -5\,\text{cm}$; 3 cm away from A towards B means $x = -5 + 3 = -2\,\text{cm}$
- Moving towards B → velocity is positive
- $a = -\omega^2(-2)$ → acceleration is positive (towards O, i.e., towards B)
- Force is positive

$$v = +,\quad a = +,\quad F = +$$

---

(f) At 4 cm away from B going towards A:
- B is at $x = +5\,\text{cm}$; 4 cm away from B towards A means $x = +5 - 4 = +1\,\text{cm}$
- Moving towards A → velocity is negative
- $a = -\omega^2(+1)$ → acceleration is negative (towards O)
- Force is negative

$$v = -,\quad a = -,\quad F = -$$

---

Summary Table:

| Case | Position | Velocity | Acceleration | Force |
|------|----------|----------|--------------|-------|
| (a) | A | Zero | + | + |
| (b) | B | Zero | − | − |
| (c) | Mid-point (→A) | − | Zero | Zero |
| (d) | 3 cm from O (→A) | − | − | − |
| (e) | 2 cm from O (→B) | + | + | + |
| (f) | 1 cm from O (→A) | − | − | − |

Condition for SHM: The acceleration must be proportional to displacement and directed opposite to it:
$$a = -\omega^2 x \quad \Rightarrow \quad a \propto -x$$

(a) $a = 0.7x$:
Here $a \propto x$ (same sign), so the acceleration is in the same direction as displacement. This does not represent SHM.

(b) $a = -200x^2$:
Here $a \propto x^2$, not $x$. The acceleration is not linearly proportional to displacement. This does not represent SHM.

(c) $a = -10x$:
Here $a \propto -x$ (opposite sign, linear). This satisfies the condition for SHM with $\omega^2 = 10$, i.e., $\omega = \sqrt{10}\,\text{rad/s}$. This represents SHM.

(d) $a = 100x^3$:
Here $a \propto x^3$, not $x$. This does not represent SHM.

Answer: Only (c) represents SHM.

Given:
- $x(t) = A\cos(\omega t + \phi)$
- At $t = 0$: $x(0) = 1\,\text{cm}$, $\dot{x}(0) = \omega\,\text{cm/s}$
- $\omega = \pi\,\text{s}^{-1}$

Step 1: Apply initial conditions to cosine form.

At $t = 0$:
$$x(0) = A\cos\phi = 1 \quad \cdots (1)$$

Velocity: $v(t) = -A\omega\sin(\omega t + \phi)$

At $t = 0$:
$$v(0) = -A\omega\sin\phi = \omega$$
$$\Rightarrow -A\sin\phi = 1 \quad \cdots (2)$$

Step 2: Find amplitude A.

Squaring and adding (1) and (2):
$$A^2\cos^2\phi + A^2\sin^2\phi = 1^2 + (-1)^2 = 2$$
$$A^2 = 2$$
$$\boxed{A = \sqrt{2}\,\text{cm}}$$

Step 3: Find phase angle φ.

Dividing (2) by (1):
$$\frac{-A\sin\phi}{A\cos\phi} = \frac{1}{1}$$
$$-\tan\phi = 1 \Rightarrow \tan\phi = -1$$

From (1): \cos\phi = \dfrac{1}{A} = \dfrac{1}{\sqrt{2}} > 0

From (2): \sin\phi = -\dfrac{1}{A} = -\dfrac{1}{\sqrt{2}} < 0

So $\phi$ is in the fourth quadrant:
$$\boxed{\phi = -\frac{\pi}{4}\,\text{rad}}$$

---

Sine function form: $x = B\sin(\omega t + \alpha)$

At $t = 0$:
$$x(0) = B\sin\alpha = 1 \quad \cdots (3)$$

Velocity: $v(t) = B\omega\cos(\omega t + \alpha)$

At $t = 0$:
$$v(0) = B\omega\cos\alpha = \omega$$
$$\Rightarrow B\cos\alpha = 1 \quad \cdots (4)$$

Amplitude B:
$$B^2\sin^2\alpha + B^2\cos^2\alpha = 1 + 1 = 2$$
$$\boxed{B = \sqrt{2}\,\text{cm}}$$

Phase α:
$$\tan\alpha = \frac{B\sin\alpha}{B\cos\alpha} = \frac{1}{1} = 1$$

From (3) and (4): both \sin\alpha > 0 and \cos\alpha > 0, so $\alpha$ is in the first quadrant:
$$\boxed{\alpha = \frac{\pi}{4}\,\text{rad}}$$

Conclusion: In both cases, the amplitude is $\sqrt{2}\,\text{cm}$. The initial phase is $-\pi/4$ for the cosine form and $\pi/4$ for the sine form.

Given:
- Maximum load on scale: $M = 50\,\text{kg}$
- Length of scale (maximum extension): $x_{\max} = 20\,\text{cm} = 0.20\,\text{m}$
- Period of oscillation: $T = 0.6\,\text{s}$
- $g = 9.8\,\text{m/s}^2$

Step 1: Find the spring constant k.

At maximum load, the spring extends by $x_{\max}$:
$$k = \frac{Mg}{x_{\max}} = \frac{50 \times 9.8}{0.20} = \frac{490}{0.20} = 2450\,\text{N/m}$$

Step 2: Find the mass m of the body using the period formula.

$$T = 2\pi\sqrt{\frac{m}{k}}$$
$$0.6 = 2\pi\sqrt{\frac{m}{2450}}$$
$$\frac{0.6}{2\pi} = \sqrt{\frac{m}{2450}}$$
$$\left(\frac{0.6}{2\pi}\right)^2 = \frac{m}{2450}$$
$$m = 2450 \times \left(\frac{0.6}{2\pi}\right)^2$$
$$m = 2450 \times \frac{0.36}{4\pi^2}$$
$$m = 2450 \times \frac{0.36}{39.478}$$
$$m = 2450 \times 0.009119$$
$$m \approx 22.36\,\text{kg}$$

Step 3: Find the weight.
$$W = mg = 22.36 \times 9.8 \approx 219\,\text{N}$$

$$\boxed{W \approx 219\,\text{N}}$$

The weight of the body is approximately 219 N (mass ≈ 22.4 kg).

Given:
- Spring constant: $k = 1200\,\text{N/m}$
- Mass: $m = 3\,\text{kg}$
- Amplitude: $A = 2.0\,\text{cm} = 0.02\,\text{m}$

Step 1: Angular frequency.
$$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{1200}{3}} = \sqrt{400} = 20\,\text{rad/s}$$

(i) Frequency of oscillations:
$$v = \frac{\omega}{2\pi} = \frac{20}{2\pi} = \frac{10}{\pi} \approx 3.18\,\text{Hz}$$

$$\boxed{v \approx 3.18\,\text{Hz}}$$

(ii) Maximum acceleration:

Maximum acceleration occurs at the extreme position ($x = A$):
$$a_{\max} = \omega^2 A = (20)^2 \times 0.02 = 400 \times 0.02 = 8\,\text{m/s}^2$$

$$\boxed{a_{\max} = 8\,\text{m/s}^2}$$

(iii) Maximum speed:

Maximum speed occurs at the mean position:
$$v_{\max} = \omega A = 20 \times 0.02 = 0.4\,\text{m/s}$$

$$\boxed{v_{\max} = 0.4\,\text{m/s}}$$

From Exercise 13.9: $\omega = 20\,\text{rad/s}$, $A = 2.0\,\text{cm} = 0.02\,\text{m}$

General equation: $x(t) = A\cos(\omega t + \phi)$

---

(a) At $t = 0$, mass is at mean position ($x = 0$) and moving in positive direction:

At $t = 0$: $x = 0 \Rightarrow A\cos\phi = 0 \Rightarrow \phi = \pm\pi/2$

Velocity at $t = 0$: v = -A\omega\sin\phi > 0 (moving in positive direction)
\Rightarrow \sin\phi < 0 \Rightarrow \phi = -\pi/2

$$\boxed{x(t) = 0.02\cos\left(20t - \frac{\pi}{2}\right) = 0.02\sin(20t)\,\text{m}}$$

---

(b) At $t = 0$, mass is at maximum stretched position ($x = +A$):

At $t = 0$: $x = +A \Rightarrow A\cos\phi = A \Rightarrow \phi = 0$

$$\boxed{x(t) = 0.02\cos(20t)\,\text{m}}$$

---

(c) At $t = 0$, mass is at maximum compressed position ($x = -A$):

At $t = 0$: $x = -A \Rightarrow A\cos\phi = -A \Rightarrow \phi = \pi$

$$\boxed{x(t) = 0.02\cos(20t + \pi) = -0.02\cos(20t)\,\text{m}}$$

---

Comparison:

All three functions have the same frequency ($\omega = 20\,\text{rad/s}$, $v \approx 3.18\,\text{Hz}$) and the same amplitude ($A = 0.02\,\text{m}$). They differ only in the initial phase $\phi$:
- Case (a): $\phi = -\pi/2$
- Case (b): $\phi = 0$
- Case (c): $\phi = \pi$

Concept: The x-projection of a particle undergoing uniform circular motion of radius $A$ and angular speed $\omega$ gives SHM: $x(t) = A\cos(\omega t + \phi_0)$, where $\phi_0$ is the initial angle of the radius vector with the positive x-axis.

Note: The exact figures are not visible, but based on the standard NCERT problem, the two cases are:

---

(a) Figure (a): Radius = 3 cm, Period $T = 2\,\text{s}$, initial position at $x = -3\,\text{cm}$ (i.e., at the leftmost point, angle = $\pi$ from positive x-axis), anticlockwise rotation.

$$\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi\,\text{rad/s}$$

Initial angle $\phi_0 = \pi$ (particle starts at $(-3, 0)$)

$$x(t) = 3\cos(\pi t + \pi) = -3\cos(\pi t)\,\text{cm}$$

$$\boxed{x(t) = -3\cos(\pi t)\,\text{cm}}$$

---

(b) Figure (b): Radius = 2 cm, Period $T = 4\,\text{s}$, initial position at $x = 0$, $y = +2\,\text{cm}$ (i.e., at the top, angle = $\pi/2$ from positive x-axis), clockwise rotation.

For clockwise rotation, the angle decreases with time:
$$\omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2}\,\text{rad/s}$$

Initial angle $\phi_0 = \pi/2$; for clockwise sense, $\phi_0 = +\pi/2$:

$$x(t) = 2\cos\left(\frac{\pi}{2}t + \frac{\pi}{2}\right) = -2\sin\left(\frac{\pi}{2}t\right)\,\text{cm}$$

$$\boxed{x(t) = -2\sin\left(\frac{\pi}{2}t\right)\,\text{cm}}$$

Concept: For SHM $x = A\cos(\omega t + \phi)$, the reference circle has:
- Radius = $A$ (amplitude)
- Angular speed = $\omega$
- Initial position of particle P on circle at angle $\phi$ from positive x-axis (anticlockwise)

First, convert each expression to the cosine form $x = A\cos(\omega t + \phi)$.

---

(a) $x = -2\sin(3t + \pi/3)$

Using $-\sin\theta = \cos(\theta + \pi/2)$:
$$x = 2\cos\left(3t + \frac{\pi}{3} + \frac{\pi}{2}\right) = 2\cos\left(3t + \frac{5\pi}{6}\right)$$

- Radius (Amplitude): $A = 2\,\text{cm}$
- Angular speed: $\omega = 3\,\text{rad/s}$
- Initial phase: $\phi = \dfrac{5\pi}{6}$ (150° from positive x-axis, anticlockwise)
- Initial position of P: at angle $150°$ from positive x-axis on the circle of radius 2 cm.

---

(b) $x = \cos(\pi/6 - t)$

Using $\cos(-\theta) = \cos\theta$:
$$x = \cos\left(t - \frac{\pi}{6}\right) = \cos\left(t + \left(-\frac{\pi}{6}\right)\right)$$

- Radius (Amplitude): $A = 1\,\text{cm}$
- Angular speed: $\omega = 1\,\text{rad/s}$
- Initial phase: $\phi = -\dfrac{\pi}{6}$ (i.e., $-30°$ or equivalently $330°$ from positive x-axis)
- Initial position of P: at angle $-30°$ (or $330°$) from positive x-axis on circle of radius 1 cm.

---

(c) $x = 3\sin(2\pi t + \pi/4)$

Using $\sin\theta = \cos(\theta - \pi/2)$:
$$x = 3\cos\left(2\pi t + \frac{\pi}{4} - \frac{\pi}{2}\right) = 3\cos\left(2\pi t - \frac{\pi}{4}\right)$$

- Radius (Amplitude): $A = 3\,\text{cm}$
- Angular speed: $\omega = 2\pi\,\text{rad/s}$
- Initial phase: $\phi = -\dfrac{\pi}{4}$ (i.e., $-45°$ or $315°$ from positive x-axis)
- Initial position of P: at angle $-45°$ from positive x-axis on circle of radius 3 cm.

---

(d) $x = 2\cos\pi t$

Already in standard cosine form with $\phi = 0$.

- Radius (Amplitude): $A = 2\,\text{cm}$
- Angular speed: $\omega = \pi\,\text{rad/s}$
- Initial phase: $\phi = 0$ (particle starts at positive x-axis)
- Initial position of P: at $(2, 0)$ on the circle of radius 2 cm.

---

Summary Table:

| Part | Amplitude (cm) | $\omega$ (rad/s) | Initial phase $\phi$ |
|------|---------------|-----------------|----------------------|
| (a) | 2 | 3 | $5\pi/6$ (150°) |
| (b) | 1 | 1 | $-\pi/6$ (−30°) |
| (c) | 3 | $2\pi$ | $-\pi/4$ (−45°) |
| (d) | 2 | $\pi$ | 0° |

(a) Maximum extension of the spring:

Case (a) — One end fixed, one mass:
The force $F$ stretches the spring. At maximum extension, the spring force equals the applied force:
$$F = kx_a \Rightarrow x_a = \frac{F}{k}$$

$$\boxed{x_a = \frac{F}{k}}$$

Case (b) — Both ends free, mass at each end:
Each end is pulled by force $F$. The spring is stretched by force $F$ at each end. The extension of the spring is determined by the force $F$ acting across it:
$$F = kx_b \Rightarrow x_b = \frac{F}{k}$$

$$\boxed{x_b = \frac{F}{k}}$$

The maximum extension is the same in both cases: $x = F/k$.

---

(b) Period of oscillation after release:

Case (a) — Mass m on spring of constant k:
$$T_a = 2\pi\sqrt{\frac{m}{k}}$$

$$\boxed{T_a = 2\pi\sqrt{\frac{m}{k}}}$$

Case (b) — Two masses m on same spring:
By symmetry, the centre of the spring does not move (it acts as a fixed point). Each half of the spring has length $L/2$ and spring constant $2k$ (since spring constant is inversely proportional to length for a uniform spring). Each mass $m$ oscillates against a spring of effective constant $2k$.

Alternatively, using the concept of reduced mass: for two equal masses $m$ connected by a spring of constant $k$, the reduced mass is:
$$\mu = \frac{m \cdot m}{m + m} = \frac{m}{2}$$

$$T_b = 2\pi\sqrt{\frac{\mu}{k}} = 2\pi\sqrt{\frac{m/2}{k}} = 2\pi\sqrt{\frac{m}{2k}}$$

$$\boxed{T_b = 2\pi\sqrt{\frac{m}{2k}}}$$

Note: $T_b = T_a/\sqrt{2}$, so the period in case (b) is smaller than in case (a).

Given:
- Stroke = $2A = 1.0\,\text{m}$ $\Rightarrow$ Amplitude $A = 0.5\,\text{m}$
- Angular frequency: $\omega = 200\,\text{rad/min} = \dfrac{200}{60}\,\text{rad/s} = \dfrac{10}{3}\,\text{rad/s}$

Formula for maximum speed in SHM:
$$v_{\max} = \omega A$$

Calculation:
$$v_{\max} = \frac{10}{3} \times 0.5 = \frac{5}{3} \approx 1.67\,\text{m/s}$$

$$\boxed{v_{\max} = \frac{10}{3}\,\text{m/s} \approx 1.67\,\text{m/s}}$$

Given:
- $g_{\text{moon}} = 1.7\,\text{m/s}^2$
- $g_{\text{earth}} = 9.8\,\text{m/s}^2$
- $T_{\text{earth}} = 3.5\,\text{s}$

Formula: $T = 2\pi\sqrt{\dfrac{L}{g}}$

For the same pendulum (same length $L$):
$$\frac{T_{\text{moon}}}{T_{\text{earth}}} = \sqrt{\frac{g_{\text{earth}}}{g_{\text{moon}}}}$$

$$T_{\text{moon}} = T_{\text{earth}} \times \sqrt{\frac{g_{\text{earth}}}{g_{\text{moon}}}}$$

$$T_{\text{moon}} = 3.5 \times \sqrt{\frac{9.8}{1.7}}$$

$$T_{\text{moon}} = 3.5 \times \sqrt{5.765}$$

$$T_{\text{moon}} = 3.5 \times 2.401$$

$$\boxed{T_{\text{moon}} \approx 8.4\,\text{s}}$$

Concept: When the car moves on a circular track with speed $v$ and radius $R$, the bob experiences a centripetal acceleration directed radially inward (horizontally) in addition to the gravitational acceleration $g$ (downward).

Effective acceleration:

The centripetal acceleration is:
$$a_c = \frac{v^2}{R}\quad\text{(horizontal, radially inward)}$$

The gravitational acceleration is:
$$g\quad\text{(vertical, downward)}$$

These two accelerations are perpendicular to each other. The effective gravitational acceleration is:
$$g_{\text{eff}} = \sqrt{g^2 + \left(\frac{v^2}{R}\right)^2}$$

Time period of the pendulum:
$$T = 2\pi\sqrt{\frac{l}{g_{\text{eff}}}}$$

$$\boxed{T = 2\pi\sqrt{\frac{l}{\sqrt{g^2 + \dfrac{v^4}{R^2}}}}}$$

Setup:
Let the cork be in equilibrium, floating in the liquid. Let $\rho$ be the density of cork and $\rho_1$ be the density of the liquid.

Equilibrium condition:
At equilibrium, the weight of the cork equals the buoyant force:
$$\rho A h g = \rho_1 A x_0 g$$
where $x_0$ is the equilibrium depth of immersion.
$$x_0 = \frac{\rho h}{\rho_1} \quad \cdots (1)$$

When displaced by a small distance $y$ downward:

The new depth of immersion = $x_0 + y$

Buoyant force = $\rho_1 A (x_0 + y) g$

Weight of cork = $\rho A h g$ (constant)

Net restoring force (upward positive):
$$F = \text{Buoyant force} - \text{Weight}$$
$$F = \rho_1 A(x_0 + y)g - \rho A h g$$
$$F = \rho_1 A x_0 g + \rho_1 A y g - \rho A h g$$

Using equilibrium condition $\rho_1 A x_0 g = \rho A h g$:
$$F = \rho A h g + \rho_1 A y g - \rho A h g$$
$$F = -\rho_1 A g y$$

(The negative sign indicates the force is opposite to displacement $y$, i.e., restoring.)

Equation of motion:
Mass of cork $m = \rho A h$
$$m\ddot{y} = -\rho_1 A g y$$
$$\rho A h \ddot{y} = -\rho_1 A g y$$
$$\ddot{y} = -\frac{\rho_1 g}{\rho h} y$$

This is of the form $\ddot{y} = -\omega^2 y$, confirming SHM with:
$$\omega^2 = \frac{\rho_1 g}{\rho h}$$

Period:
$$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{\rho h}{\rho_1 g}}$$

$$\boxed{T = 2\pi\sqrt{\frac{h\rho}{\rho_1 g}}}$$

Hence proved.

Setup:
Let the U-tube have uniform cross-sectional area $A$. Let the total length of mercury in the U-tube be $L$. When the suction pump is removed, let the mercury be displaced by a small distance $y$ from its equilibrium position.

Restoring force analysis:

When mercury in one arm is displaced by $y$ downward (and rises by $y$ in the other arm), the difference in height of mercury columns between the two arms is $2y$.

The restoring force due to this height difference is the weight of the extra mercury column of height $2y$:
$$F = -\rho A (2y) g$$

where $\rho$ is the density of mercury.

The negative sign indicates the force is directed opposite to the displacement (restoring).

Equation of motion:

Total mass of mercury: $m = \rho A L$

$$m\ddot{y} = -2\rho A g y$$
$$\rho A L \ddot{y} = -2\rho A g y$$
$$\ddot{y} = -\frac{2g}{L} y$$

This is of the form $\ddot{y} = -\omega^2 y$ with:
$$\omega^2 = \frac{2g}{L}$$

Since the acceleration is proportional to displacement and directed opposite to it, the mercury column executes simple harmonic motion.

Period of oscillation:
$$T = 2\pi\sqrt{\frac{L}{2g}}$$

Hence, when the suction pump is removed, the mercury column in the U-tube executes SHM. $\blacksquare$