OscillationsClass 11 Physics NCERT Solutions

(a) Periodic motion: The swimmer repeats the motion of swimming from one bank to the other and back at regular time intervals.

(b) Periodic motion: A freely suspended magnet, when displaced, oscillates about its mean N-S direction. This to-and-fro motion repeats at regular intervals, so it is periodic.

(c) Periodic motion: The rotation of a hydrogen molecule about its centre of mass is a repetitive motion that occurs at a regular interval, hence it is periodic.

(d) Non-periodic motion: An arrow released from a bow moves in a projectile path. It does not repeat its motion, so it is non-periodic.

(a) Periodic but not simple harmonic motion: The rotation of the Earth is periodic because it repeats every 24 hours. However, it is not an oscillatory (to-and-fro) motion about a mean position, so it is not SHM.

(b) Simple harmonic motion (nearly): For small displacements, the restoring force on the mercury column is directly proportional to the displacement from the equilibrium position. Therefore, the motion is approximately simple harmonic.

(c) Simple harmonic motion (nearly): When a ball bearing is released slightly above the lowest point of a smooth curved bowl, the restoring force is proportional to the sine of the angular displacement ($mg \sin \theta$). For small angles, $\sin \theta \approx \theta$, making the restoring force proportional to the displacement. Thus, the motion is nearly simple harmonic.

(d) Periodic but not simple harmonic motion: The general vibrations of a polyatomic molecule are complex. They are a superposition of several simple harmonic motions of different frequencies. The resultant motion is periodic but not simple harmonic.

A motion is periodic if it repeats itself after a fixed interval of time.

(a) Non-periodic: The plot shows the position $x$ increasing linearly with time $t$. The motion does not repeat itself.

(b) Periodic: The plot shows that the motion repeats itself after every 2 seconds. The period of motion is $T = 2 \text{ s}$.

(c) Non-periodic: The plot shows an oscillatory motion with decreasing amplitude. Since the motion does not repeat itself identically, it is non-periodic. This represents damped oscillations.

(d) Periodic: The plot shows a motion that repeats itself after every 2 seconds. The period of motion is $T = 2 \text{ s}$.

(a) Simple harmonic motion (SHM): The function is $f(t) = \sin \omega t - \cos \omega t$. This can be written as: $f(t) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \omega t - \frac{1}{\sqrt{2}} \cos \omega t \right)$ $f(t) = \sqrt{2} (\cos(\frac{\pi}{4}) \sin \omega t - \sin(\frac{\pi}{4}) \cos \omega t)$ $f(t) = \sqrt{2} \sin(\omega t - \frac{\pi}{4})$ This is the equation of SHM. The period is $T = \frac{2\pi}{\omega}$.

(b) Periodic but not simple harmonic motion: The function is $f(t) = \sin^3 \omega t$. Using the trigonometric identity $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$, we get: $\sin^3 \omega t = \frac{1}{4} (3\sin \omega t - \sin 3\omega t)$ This function is a superposition of two SHMs with different angular frequencies ($\omega$ and $3\omega$). Therefore, it is periodic but not simple harmonic. The period of $\sin \omega t$ is $T_1 = 2\pi/\omega$. The period of $\sin 3\omega t$ is $T_2 = 2\pi/(3\omega)$. The period of the combined function is the least common multiple of $T_1$ and $T_2$, which is $T = \frac{2\pi}{\omega}$.

(c) Simple harmonic motion (SHM): The function is $f(t) = 3 \cos (\pi/4 - 2\omega t) = 3 \cos (2\omega t - \pi/4)$ (since $\cos(-\theta) = \cos\theta$). This represents SHM with an angular frequency of $2\omega$. The period is $T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$.

(d) Periodic but not simple harmonic motion: The function is $f(t) = \cos \omega t + \cos 3\omega t + \cos 5\omega t$. This is a superposition of three SHMs with different angular frequencies ($\omega, 3\omega, 5\omega$). The resultant motion is periodic but not SHM. The period is the least common multiple of the individual periods ($2\pi/\omega, 2\pi/(3\omega), 2\pi/(5\omega)$), which is $T = \frac{2\pi}{\omega}$.

(e) Non-periodic motion: The function $f(t) = \exp(-\omega^2 t^2)$ decreases from 1 (at $t=0$) towards 0 as $t$ increases. It does not repeat its value, so it is non-periodic.

(f) Non-periodic motion: The function $f(t) = 1 + \omega t + \omega^2 t^2$ increases with time and does not repeat its value. It is a non-periodic function.

Let the midpoint of AB be the origin ($x=0$). The distance AB is 10 cm, so the amplitude is $A = 5$ cm. Point A is at $x = -5$ cm and point B is at $x = +5$ cm. The direction from A to B is positive. Velocity ($v$) is positive when moving towards B and negative when moving towards A. Acceleration ($a$) and Force ($F$) are given by $a = -\omega^2 x$ and $F = -kx$. So, their sign is opposite to the sign of displacement ($x$).

(a) At the end A: Position: $x = -5$ cm (negative). Velocity: $v = 0$. Acceleration: $a$ is positive (opposite to $x$). Force: $F$ is positive (opposite to $x$).

(b) At the end B: Position: $x = +5$ cm (positive). Velocity: $v = 0$. Acceleration: $a$ is negative. Force: $F$ is negative.

(c) At the mid-point of AB going towards A: Position: $x = 0$. Velocity: $v$ is negative (moving towards A). Acceleration: $a = 0$. Force: $F = 0$.

(d) At 2 cm away from B going towards A: Position: $x = +5 - 2 = +3$ cm (positive). Velocity: $v$ is negative (moving towards A). Acceleration: $a$ is negative. Force: $F$ is negative.

(e) At 3 cm away from A going towards B: Position: $x = -5 + 3 = -2$ cm (negative). Velocity: $v$ is positive (moving towards B). Acceleration: $a$ is positive. Force: $F$ is positive.

(f) At 4 cm away from B going towards A: Position: $x = +5 - 4 = +1$ cm (positive). Velocity: $v$ is negative (moving towards A). Acceleration: $a$ is negative. Force: $F$ is negative.

The condition for simple harmonic motion (SHM) is that the acceleration ($a$) must be directly proportional to the displacement ($x$) from the equilibrium position and directed opposite to the displacement. Mathematically, this is expressed as: $a = -\omega^2 x$ where $\omega^2$ is a positive constant.

Let's analyze the given relationships: (a) $a = 0.7x$: Here, acceleration is proportional to displacement, but it is in the same direction as the displacement (not opposite). This does not represent SHM.

(b) $a = -200x^2$: Here, acceleration is proportional to the square of the displacement ($x^2$), not to $x$. This does not represent SHM.

(c) $a = -10x$: This relationship fits the form $a = -\omega^2 x$, with $\omega^2 = 10$. The acceleration is proportional to the displacement and is in the opposite direction. This represents SHM.

(d) $a = 100x^3$: Here, acceleration is proportional to the cube of the displacement ($x^3$), not to $x$. This does not represent SHM.

Final Answer: The relationship $a = -10x$ involves simple harmonic motion.

Case 1: Cosine function Given: Displacement function: $x(t) = A \cos(\omega t + \phi)$ Initial position ($t=0$): $x(0) = 1 \text{ cm}$ Initial velocity ($t=0$): $v(0) = \omega \text{ cm/s}$ Angular frequency: $\omega = \pi \text{ s}^{-1}$

Calculation: At $t=0$, $x(0) = A \cos(\phi) = 1$ ---(1)

The velocity function is $v(t) = \frac{dx}{dt} = -A\omega \sin(\omega t + \phi)$. At $t=0$, $v(0) = -A\omega \sin(\phi) = \omega$ This simplifies to $-A \sin(\phi) = 1$ ---(2)

To find the amplitude $A$, we square and add equations (1) and (2): $(A \cos \phi)^2 + (-A \sin \phi)^2 = 1^2 + 1^2$ $A^2 (\cos^2 \phi + \sin^2 \phi) = 2$ $A^2 = 2 \implies A = \sqrt{2} \text{ cm}$

To find the initial phase $\phi$, we divide equation (2) by equation (1): $\frac{-A \sin \phi}{A \cos \phi} = \frac{1}{1}$ $\tan \phi = -1$ From (1), $\cos \phi = 1/A = 1/\sqrt{2} > 0$. From (2), $\sin \phi = -1/A = -1/\sqrt{2} < 0$. Since $\cos \phi$ is positive and $\sin \phi$ is negative, $\phi$ must be in the fourth quadrant. Therefore, $\phi = -\frac{\pi}{4}$ or $\frac{7\pi}{4}$ radians.

Case 2: Sine function Given: Displacement function: $x(t) = B \sin(\omega t + \alpha)$ Initial conditions are the same.

Calculation: At $t=0$, $x(0) = B \sin(\alpha) = 1$ ---(3)

The velocity function is $v(t) = \frac{dx}{dt} = B\omega \cos(\omega t + \alpha)$. At $t=0$, $v(0) = B\omega \cos(\alpha) = \omega$ This simplifies to $B \cos(\alpha) = 1$ ---(4)

To find the amplitude $B$, we square and add equations (3) and (4): $(B \sin \alpha)^2 + (B \cos \alpha)^2 = 1^2 + 1^2$ $B^2 (\sin^2 \alpha + \cos^2 \alpha) = 2$ $B^2 = 2 \implies B = \sqrt{2} \text{ cm}$

To find the initial phase $\alpha$, we divide equation (3) by equation (4): $\frac{B \sin \alpha}{B \cos \alpha} = \frac{1}{1}$ $\tan \alpha = 1$ From (3), $\sin \alpha = 1/B = 1/\sqrt{2} > 0$. From (4), $\cos \alpha = 1/B = 1/\sqrt{2} > 0$. Since both $\sin \alpha$ and $\cos \alpha$ are positive, $\alpha$ must be in the first quadrant. Therefore, $\alpha = \frac{\pi}{4}$ radians.

Final Answer: For $x(t) = A \cos(\omega t + \phi)$: Amplitude $A = \sqrt{2} \text{ cm}$, Initial phase $\phi = -\pi/4$ rad. For $x(t) = B \sin(\omega t + \alpha)$: Amplitude $B = \sqrt{2} \text{ cm}$, Initial phase $\alpha = \pi/4$ rad.

Given: Maximum mass, $M_{max} = 50 \text{ kg}$ Length of the scale (maximum extension), $x_{max} = 20 \text{ cm} = 0.2 \text{ m}$ Period of oscillation, $T = 0.6 \text{ s}$ Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$

To Find: The weight of the body, $W$.

Calculation: Step 1: Find the spring constant (k). The maximum force the spring can measure corresponds to the weight of 50 kg. $F_{max} = M_{max} \times g = 50 \text{ kg} \times 9.8 \text{ m/s}^2 = 490 \text{ N}$ This force causes an extension of $x_{max} = 0.2 \text{ m}$. $k = \frac{F_{max}}{x_{max}} = \frac{490 \text{ N}}{0.2 \text{ m}} = 2450 \text{ N/m}$

Step 2: Find the mass of the body (m). The body oscillates with a period $T = 0.6 \text{ s}$. We can use the formula for the period to find its mass. $T = 2\pi \sqrt{\frac{m}{k}}$ Squaring both sides: $T^2 = (2\pi)^2 \frac{m}{k}$ Rearranging for $m$: $m = \frac{k T^2}{4\pi^2} = \frac{(2450 \text{ N/m}) \times (0.6 \text{ s})^2}{4 \times (3.1416)^2}$ $m = \frac{2450 \times 0.36}{39.48} \approx 22.36 \text{ kg}$

Step 3: Find the weight of the body (W). $W = m \times g = 22.36 \text{ kg} \times 9.8 \text{ m/s}^2 \approx 219.1 \text{ N}$

Final Answer: The weight of the body is approximately $219.1 \text{ N}$.

From Exercise 13.9, we have: Amplitude, $A = 0.02 \text{ m}$ Angular frequency, $\omega = 20 \text{ rad/s}$

The general equation for displacement in SHM can be written as $x(t) = A \cos(\omega t + \phi)$, where $\phi$ is the initial phase.

(a) At the mean position ($x=0$) at $t=0$: If the mass is at the mean position, its displacement is zero. However, it must be moving. Let's assume it is moving towards the positive direction. The velocity is maximum and positive. Using $x(t) = A \sin(\omega t + \phi')$ form is easier here. At $t=0$, $x(0) = A \sin(\phi') = 0$, which means $\phi'=0$ or $\pi$. Since velocity $v(t) = A\omega \cos(\omega t + \phi')$ is positive at $t=0$, $A\omega \cos(\phi') > 0$, so $\cos(\phi') > 0$. This implies $\phi'=0$. So, the equation is $x(t) = A \sin(\omega t) = 0.02 \sin(20t)$. (If it were moving in the negative direction, the equation would be $x(t) = -0.02 \sin(20t)$).

(b) At the maximum stretched position ($x=+A$) at $t=0$: Here, the initial displacement is maximum and positive. This corresponds to the cosine function with zero initial phase. $x(0) = A \cos(\phi) = A \implies \cos(\phi) = 1 \implies \phi = 0$. So, the equation is $x(t) = A \cos(\omega t) = 0.02 \cos(20t)$.

(c) At the maximum compressed position ($x=-A$) at $t=0$: Here, the initial displacement is maximum and negative. $x(0) = A \cos(\phi) = -A \implies \cos(\phi) = -1 \implies \phi = \pi$. So, the equation is $x(t) = A \cos(\omega t + \pi) = -A \cos(\omega t) = -0.02 \cos(20t)$.

Difference between the functions: The three functions for SHM have the same amplitude ($A = 0.02 \text{ m}$) and the same frequency (or angular frequency $\omega = 20 \text{ rad/s}$). They differ only in their initial phase.

The $x$-projection of the radius vector of a particle in uniform circular motion gives the simple harmonic motion described by the equation: $x(t) = A \cos(\omega t + \phi)$ where $A$ is the radius, $\omega = 2\pi/T$ is the angular speed, and $\phi$ is the initial phase angle (the angle at $t=0$ with the positive x-axis).

(a) Figure (a): Given: Radius, $A = 3 \text{ cm}$ Period, $T = 2 \text{ s}$ Sense of revolution: Anti-clockwise Initial position ($t=0$): Particle P is at $(3, 0)$ on the positive x-axis.

Calculation: Angular speed: $\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi \text{ rad/s}$. Initial phase: At $t=0$, the particle is on the positive x-axis, so the initial angle with the positive x-axis is $\phi = 0$ radians. The motion is anti-clockwise, which is the standard positive direction for angle measurement.

The equation for the x-projection is: $x(t) = A \cos(\omega t + \phi) = 3 \cos(\pi t + 0)$ $x(t) = 3 \cos(\pi t) \text{ cm}$

(b) Figure (b): Given: Radius, $B = 2 \text{ m}$ Period, $T = 4 \text{ s}$ Sense of revolution: Clockwise Initial position ($t=0$): Particle P is at $(2, 0)$ on the positive x-axis.

Calculation: Angular speed: $\omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2} \text{ rad/s}$. Initial phase: At $t=0$, the particle is on the positive x-axis, so the initial angle is $\phi = 0$ radians. Since the motion is clockwise, the angle at time $t$ will be $-\omega t$.

The equation for the x-projection is: $x(t) = B \cos(-\omega t + \phi) = 2 \cos(-\frac{\pi}{2} t + 0)$ Since $\cos(-\theta) = \cos(\theta)$, we have: $x(t) = 2 \cos(\frac{\pi}{2} t) \text{ m}$

Final Answer: (a) $x(t) = 3 \cos(\pi t)$ cm (b) $x(t) = 2 \cos(\frac{\pi}{2} t)$ m

We will convert each equation to the standard form $x(t) = A \cos(\omega t + \phi)$ for an anti-clockwise rotation.

(a) $x = -2 \sin(3t + \pi/3)$ Using $\sin(\theta) = \cos(\theta - \pi/2)$ and $-\cos(\theta) = \cos(\theta + \pi)$: $x = -2 \cos(3t + \pi/3 - \pi/2) = -2 \cos(3t - \pi/6)$ $x = 2 \cos(3t - \pi/6 + \pi) = 2 \cos(3t + 5\pi/6)$

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

(c) $x = 3 \sin(2\pi t + \pi/4)$ Using $\sin(\theta) = \cos(\theta - \pi/2)$: $x = 3 \cos(2\pi t + \pi/4 - \pi/2) = 3 \cos(2\pi t - \pi/4)$

(d) $x = 2 \cos \pi t$ This is already in standard form.

Plots of Reference Circles: (A description of the plots is provided as images cannot be generated)

(a) Draw a circle with radius 2 cm centered at the origin. The initial position vector (at $t=0$) is drawn at an angle of $5\pi/6$ (150 degrees) anti-clockwise from the positive x-axis. An arrow indicates anti-clockwise rotation with $\omega = 3$ rad/s.

(b) Draw a circle with radius 1 cm. The initial position vector is at an angle of $-\pi/6$ (-30 degrees) or $11\pi/6$ (330 degrees) from the positive x-axis. An arrow indicates anti-clockwise rotation with $\omega = 1$ rad/s.

(c) Draw a circle with radius 3 cm. The initial position vector is at an angle of $-\pi/4$ (-45 degrees) or $7\pi/4$ (315 degrees) from the positive x-axis. An arrow indicates anti-clockwise rotation with $\omega = 2\pi$ rad/s.

(d) Draw a circle with radius 2 cm. The initial position vector is along the positive x-axis (angle 0). An arrow indicates anti-clockwise rotation with $\omega = \pi$ rad/s.

Case (a): A force $\mathbf{F}$ is applied to the free end of the spring, which is fixed at the other end. According to Hooke's law, the extension $x_a$ is given by: $F = k x_a \implies x_a = \frac{F}{k}$

Case (b): The spring is pulled by a force $\mathbf{F}$ at both ends. This situation is equivalent to fixing the center of the spring and pulling each half with a force $\mathbf{F}$. Alternatively, and more simply, the tension throughout the spring is uniform and equal to $F$. The situation is mechanically identical to case (a), where the clamp provides the reaction force $\mathbf{F}$. Therefore, the total extension $x_b$ is: $x_b = \frac{F}{k}$ Thus, the maximum extension is the same in both cases.

Case (a): A single mass $m$ is attached to a spring of constant $k$. When released, it performs SHM. The period of oscillation $T_a$ is given by the standard formula: $T_a = 2\pi \sqrt{\frac{m}{k}}$

Case (b): Two masses, each of mass $m$, are attached to the ends of the spring. The system will oscillate symmetrically about its center of mass, which remains stationary. This system is equivalent to two separate systems, each with a mass $m$ attached to a spring of half the original length. A spring of half the length has a spring constant of $2k$. So, for one mass, the period would be $T = 2\pi \sqrt{\frac{m}{2k}}$.

Alternatively, we can analyze this as a two-body problem using the concept of reduced mass, $\mu$. $\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{m \times m}{m + m} = \frac{m^2}{2m} = \frac{m}{2}$ The period of oscillation of the separation between the two masses is given by: $T_b = 2\pi \sqrt{\frac{\mu}{k}} = 2\pi \sqrt{\frac{m/2}{k}} = 2\pi \sqrt{\frac{m}{2k}}$

Final Answer: (a) The maximum extension is $F/k$ in both cases. (b) The period of oscillation in case (a) is $T_a = 2\pi \sqrt{\frac{m}{k}}$. The period of oscillation in case (b) is $T_b = 2\pi \sqrt{\frac{m}{2k}}$.

Given: Stroke = $1.0 \text{ m}$ Angular frequency, $\omega = 200 \text{ rad/min}$

To Find: Maximum speed, $v_{max}$

Formula: The maximum speed in SHM is given by $v_{max} = A\omega$, where $A$ is the amplitude.

Calculation: Step 1: Find the amplitude (A). The stroke is twice the amplitude. Stroke = $2A = 1.0 \text{ m}$ $A = \frac{1.0}{2} = 0.5 \text{ m}$

Step 2: Convert angular frequency to SI units (rad/s). $\omega = 200 \frac{\text{rad}}{\text{min}} = 200 \frac{\text{rad}}{60 \text{ s}} = \frac{20}{6} \text{ rad/s} = \frac{10}{3} \text{ rad/s}$

Step 3: Calculate the maximum speed ($v_{max}$). $v_{max} = A\omega = (0.5 \text{ m}) \times (\frac{10}{3} \text{ rad/s})$ $v_{max} = \frac{5}{3} \text{ m/s} \approx 1.67 \text{ m/s}$

Final Answer: The maximum speed of the piston is approximately $1.67 \text{ m/s}$.

Given: Acceleration due to gravity on the Moon, $g_m = 1.7 \text{ m s}^{-2}$ Acceleration due to gravity on the Earth, $g_e = 9.8 \text{ m s}^{-2}$ Time period on the Earth, $T_e = 3.5 \text{ s}$

To Find: Time period on the Moon, $T_m$

Formula: The time period of a simple pendulum is given by: $T = 2\pi \sqrt{\frac{L}{g}}$ where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity.

Calculation: For the pendulum on Earth: $T_e = 2\pi \sqrt{\frac{L}{g_e}}$ ---(1)

For the same pendulum on the Moon (length $L$ is constant): $T_m = 2\pi \sqrt{\frac{L}{g_m}}$ ---(2)

Divide equation (2) by equation (1): $\frac{T_m}{T_e} = \frac{2\pi \sqrt{L/g_m}}{2\pi \sqrt{L/g_e}} = \sqrt{\frac{L}{g_m} \times \frac{g_e}{L}} = \sqrt{\frac{g_e}{g_m}}$

Now, solve for $T_m$: $T_m = T_e \sqrt{\frac{g_e}{g_m}}$ $T_m = 3.5 \text{ s} \times \sqrt{\frac{9.8 \text{ m s}^{-2}}{1.7 \text{ m s}^{-2}}}$ $T_m = 3.5 \times \sqrt{5.765} \approx 3.5 \times 2.401$ $T_m \approx 8.4 \text{ s}$

Final Answer: The time period of the simple pendulum on the surface of the Moon is approximately $8.4 \text{ s}$.

When the car moves on a circular track, the bob of the pendulum experiences two accelerations:

The pendulum will hang at an angle to the vertical such that the tension in the string provides the necessary forces to balance gravity and provide the centripetal force. The effective acceleration, $g_{eff}$, is the vector sum of $g$ and $a_c$. Since these two accelerations are perpendicular to each other, the magnitude of the effective acceleration is: $g_{eff} = \sqrt{g^2 + a_c^2} = \sqrt{g^2 + \left(\frac{v^2}{R}\right)^2} = \sqrt{g^2 + \frac{v^4}{R^2}}$ This effective acceleration determines the restoring force for small oscillations. The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$. For this case, we replace $g$ with $g_{eff}$.

Therefore, the time period of small oscillations in the radial direction about the equilibrium position will be: $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$ $T = 2\pi \sqrt{\frac{l}{\sqrt{g^2 + v^4/R^2}}}$

Final Answer: The time period of the pendulum will be $T = 2\pi \sqrt{\frac{l}{(g^2 + v^4/R^2)^{1/2}}}$.

Step 1: Equilibrium Condition Let the cork be floating in the liquid. Let $h_0$ be the length of the cork submerged in the liquid at equilibrium. Weight of the cork = Buoyant force $W = F_B$ $(A h \rho) g = (A h_0 \rho_l) g$ $h \rho = h_0 \rho_l$ ---(1)

Step 2: Restoring Force Now, let the cork be depressed by a small distance $y$ from its equilibrium position. The new submerged length is $(h_0 + y)$. The new buoyant force is $F'_B = A(h_0 + y)\rho_l g$. The net force on the cork is the difference between the new buoyant force and its weight, directed upwards. $F_{net} = W - F'_B$ $F_{net} = (A h \rho) g - A(h_0 + y)\rho_l g$ From equation (1), we know $(A h \rho) g = (A h_0 \rho_l) g$. Substituting this into the net force equation: $F_{net} = (A h_0 \rho_l) g - A(h_0 + y)\rho_l g$ $F_{net} = A h_0 \rho_l g - A h_0 \rho_l g - A y \rho_l g$ $F_{net} = - (A \rho_l g) y$ The negative sign indicates that the net force is directed towards the equilibrium position (restoring force).

Step 3: Proving SHM and Finding the Period The net force is of the form $F = -ky$, where the effective spring constant is $k = A \rho_l g$. Since the restoring force is directly proportional to the displacement $y$ and oppositely directed, the motion is simple harmonic.

The mass of the cork is $m = \text{Volume} \times \text{density} = A h \rho$. The period of oscillation is given by: $T = 2\pi \sqrt{\frac{m}{k}}$ $T = 2\pi \sqrt{\frac{A h \rho}{A \rho_l g}}$ $T = 2\pi \sqrt{\frac{h\rho}{\rho_l g}}$ Hence, it is shown that the cork oscillates simple harmonically with the given period.

Step 1: Setup and Equilibrium Consider a U-tube of uniform cross-sectional area $A$ containing mercury of density $\rho$. Let the total length of the mercury column be $L$. In equilibrium, the mercury level is the same in both arms.

Step 2: Displacement and Restoring Force When the suction pump is removed, let's assume the mercury level in one arm is displaced downwards by a distance $y$ from the equilibrium position. To maintain the volume, the level in the other arm must rise by the same distance $y$. The difference in height between the mercury levels in the two arms is now $h = y + y = 2y$. This height difference creates a pressure difference, resulting in a restoring force. The restoring force is equal to the weight of the mercury column of height $2y$. Weight of the column = (Volume of the column) $\times$ density $\times g$ $F_{restore} = -(A \times 2y) \times \rho \times g$ $F_{restore} = -(2A\rho g) y$ The negative sign indicates that the force is directed towards the equilibrium position, opposing the displacement.

Step 3: Proving SHM The restoring force is directly proportional to the displacement $y$ ($F \propto -y$). This is the condition for simple harmonic motion. Therefore, the column of mercury in the U-tube executes simple harmonic motion.

Additional (Finding the Period): The total mass of the mercury oscillating is $m = (\text{Total Length}) \times (\text{Area}) \times (\text{density}) = L A \rho$. From the force equation, we can identify the effective spring constant $k = 2A\rho g$. The period of oscillation $T$ is given by: $T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{L A \rho}{2 A \rho g}} = 2\pi \sqrt{\frac{L}{2g}}$ This confirms the motion is SHM, as it has a well-defined period dependent on the system's physical properties.