NCERT Solutions · Class 11 Physics Oscillations

Key idea — the x-projection of uniform circular motion is SHM. If P moves round a circle of radius \(\displaystyle A\) at constant angular speed \(\displaystyle \omega\), and at time \(\displaystyle t\) the radius vector \(\displaystyle OP\) makes an angle \(\displaystyle \theta(t)\) with the \(\displaystyle +x\)-axis (measured anticlockwise), then the foot of the perpendicular from P onto the x-axis sits at\[x(t) = A\cos\theta(t), \qquad \theta(t) = \theta_0 \pm \omega t, \qquad \omega = \frac{2\pi}{T} \]Here \(\displaystyle A\) = radius of the circle = amplitude of the SHM, \(\displaystyle T\) = period of revolution = period of the SHM, \(\displaystyle \theta_0\) = angle of \(\displaystyle OP\) at \(\displaystyle t=0\), and \(\displaystyle \omega\) = angular speed. Use \(\displaystyle +\omega t\) if the particle revolves anticlockwise (the angle grows) and \(\displaystyle -\omega t\) if it revolves clockwise (the angle shrinks). So the whole job is: read \(\displaystyle A\), \(\displaystyle T\), \(\displaystyle \theta_0\) and the sense off the picture, then write the cosine.Case (a) — the small circle. Reading off the figure: the label "$\displaystyle 3$ cm" sits on the segment from the centre down to the circle, so \(\displaystyle A = 3\ \text{cm}\); "T = $\displaystyle 2$ s" is printed at the left of the x-axis, so \(\displaystyle T = 2\ \text{s}\); "P (t = $\displaystyle 0$)" is marked at the bottom of the circle, on the negative y-axis; the blue arrowhead on the lower-left arc points up and to the left, i.e. P travels from the bottom towards the \(\displaystyle -x\) side, which is clockwise.Angular speed:\[\omega = \frac{2\pi}{T} = \frac{2\pi}{2\ \text{s}} = \pi\ \text{rad s}^{-1} \]At \(\displaystyle t=0\), P is on the negative y-axis, so \(\displaystyle \theta_0 = -\dfrac{\pi}{2}\). The motion is clockwise, so the angle decreases:\[\theta(t) = -\frac{\pi}{2} - \omega t = -\frac{\pi}{2} - \pi t \]Project onto the x-axis, using \(\displaystyle \cos(-\phi)=\cos\phi\):\[x(t) = A\cos\theta(t) = 3\cos\!\left(-\frac{\pi}{2}-\pi t\right)\ \text{cm} = 3\cos\!\left(\pi t + \frac{\pi}{2}\right)\ \text{cm} \]and since \(\displaystyle \cos\!\left(\phi + \frac{\pi}{2}\right) = -\sin\phi\),\[x(t) = -3\sin(\pi t)\ \text{cm} \]Check it against the picture: at \(\displaystyle t=0\), \(\displaystyle x = 0\) — correct, P is at the bottom, directly below the centre. A quarter period later, \(\displaystyle t = 0.5\ \text{s}\), \(\displaystyle x = -3\sin(\pi/2) = -3\ \text{cm}\), i.e. P has reached the leftmost point — which is exactly where the clockwise arrow was heading.Case (b) — the large circle. Reading off the figure: "$\displaystyle 2$ m" is written along the horizontal from P to the centre, so the radius is \(\displaystyle A = 2\ \text{m}\); "T = $\displaystyle 4$ s" is printed at the right of the x-axis, so \(\displaystyle T = 4\ \text{s}\); "P (t = $\displaystyle 0$)" labels the left end of the horizontal diameter, on the negative x-axis; the blue arrowhead on the lower-left arc points down and to the right, i.e. P moves from the left point towards the bottom of the circle, which is anticlockwise.Angular speed:\[\omega = \frac{2\pi}{T} = \frac{2\pi}{4\ \text{s}} = \frac{\pi}{2}\ \text{rad s}^{-1} \]At \(\displaystyle t=0\), P is on the negative x-axis, so \(\displaystyle \theta_0 = \pi\). The motion is anticlockwise, so the angle increases:\[\theta(t) = \pi + \omega t = \pi + \frac{\pi}{2}t \]Project onto the x-axis, using \(\displaystyle \cos(\pi + \phi) = -\cos\phi\):\[x(t) = A\cos\theta(t) = 2\cos\!\left(\frac{\pi}{2}t + \pi\right)\ \text{m} = -2\cos\!\left(\frac{\pi t}{2}\right)\ \text{m} \]Check it against the picture: at \(\displaystyle t=0\), \(\displaystyle x = -2\ \text{m}\) — correct, P is at the far left. At \(\displaystyle t = 1\ \text{s}\) (a quarter period), \(\displaystyle x = -2\cos(\pi/2) = 0\), i.e. P is at the bottom of the circle, which is where the anticlockwise arrow points.Note that only the phase convention is a choice, not the physics: writing (a) as \(\displaystyle x = 3\cos\!\left(\pi t + \frac{\pi}{2}\right)\) cm or (b) as \(\displaystyle x = 2\cos\!\left(\frac{\pi t}{2} + \pi\right)\) m describes the identical motion. In both cases the amplitude is the circle's radius and the SHM period equals the period of revolution.Answer: (a) \(\displaystyle x(t) = -3\sin(\pi t)\ \text{cm}\) (amplitude $\displaystyle 3$ cm, \(\displaystyle \omega = \pi\ \text{rad s}^{-1}\)); (b) \(\displaystyle x(t) = -2\cos\!\left(\dfrac{\pi t}{2}\right)\ \text{m}\) (amplitude $\displaystyle 2$ m, \(\displaystyle \omega = \pi/2\ \text{rad s}^{-1}\)).

Every SHM is the shadow of a point going round a circle at constant speed — you just have to write \(\displaystyle x(t)\) as a cosine to read off the circle.The textbook's reference-circle picture is this: a point \(\displaystyle P\) moves anticlockwise on a circle of radius \(\displaystyle A\) with constant angular speed \(\displaystyle \omega\). If \(\displaystyle P\) starts (at \(\displaystyle t=0\)) at an angle \(\displaystyle \phi\) measured anticlockwise from the positive x-axis, then at time \(\displaystyle t\) it has turned through an extra angle \(\displaystyle \omega t\), so its angular position is \(\displaystyle (\omega t+\phi)\), and the projection of \(\displaystyle OP\) onto the x-axis is \[x(t) = A\cos(\omega t+\phi). \] So to "plot the reference circle" for a given \(\displaystyle x(t)\), force it into exactly this cosine form. Then radius \(\displaystyle =A\), angular speed \(\displaystyle =\omega\) (anticlockwise, as the question fixes), and the initial position \(\displaystyle P_0\) sits at angle \(\displaystyle \phi\) on the circle, i.e. at Cartesian point \(\displaystyle (A\cos\phi,\;A\sin\phi)\).Because the given functions are mostly written with sine, two identities do the conversion (both just shift the reference circle's starting angle — they do not change the physics): \[\sin\theta=\cos\!\left(\theta-\tfrac{\pi}{2}\right),\qquad -\sin\theta=\cos\!\left(\theta+\tfrac{\pi}{2}\right). \] A sign in front of the sine is the part people mis-handle — get it into cosine form before reading off \(\displaystyle \phi\), do not just copy the number sitting next to \(\displaystyle t\).(a) \(\displaystyle x=-2\sin(3t+\pi/3)\)Using \(\displaystyle -\sin\theta=\cos(\theta+\pi/2)\): \[x=2\cos\!\left(3t+\frac{\pi}{3}+\frac{\pi}{2}\right)=2\cos\!\left(3t+\frac{5\pi}{6}\right)\ \text{cm}. \] So \(\displaystyle A=2\ \text{cm}\), \(\displaystyle \omega=3\ \text{rad s}^{-1}\), \(\displaystyle \phi=5\pi/6\ \text{rad}=150^\circ\).Reference circle: radius \(\displaystyle 2\ \text{cm}\), centre at the origin, rotating anticlockwise at \(\displaystyle 3\ \text{rad s}^{-1}\). The particle \(\displaystyle P_0\) starts \(\displaystyle 150^\circ\) anticlockwise from the positive x-axis, i.e. at \[P_0=\left(2\cos150^\circ,\;2\sin150^\circ\right)=(-\sqrt3,\,1)\ \text{cm}\approx(-1.73,\,1.00)\ \text{cm}, \] which is in the second quadrant, just above the negative x-axis.(b) \(\displaystyle x=\cos(\pi/6-t)\)Cosine is an even function, \(\displaystyle \cos(-\theta)=\cos\theta\), so \[x=\cos\!\left(t-\frac{\pi}{6}\right)\ \text{cm}. \] This is already the standard form with \(\displaystyle A=1\ \text{cm}\), \(\displaystyle \omega=1\ \text{rad s}^{-1}\), \(\displaystyle \phi=-\pi/6\ \text{rad}=-30^\circ\).A negative \(\displaystyle \phi\) does not mean clockwise rotation — the rotation is still anticlockwise at \(\displaystyle \omega\); it only means the starting point sits below the x-axis. Reference circle: radius \(\displaystyle 1\ \text{cm}\), anticlockwise at \(\displaystyle 1\ \text{rad s}^{-1}\), starting point \[P_0=\left(\cos(-30^\circ),\sin(-30^\circ)\right)=\left(\frac{\sqrt3}{2},-\frac12\right)\ \text{cm}\approx(0.866,-0.500)\ \text{cm}, \] in the fourth quadrant, \(\displaystyle 30^\circ\) below the positive x-axis.(c) \(\displaystyle x=3\sin(2\pi t+\pi/4)\)Using \(\displaystyle \sin\theta=\cos(\theta-\pi/2)\): \[x=3\cos\!\left(2\pi t+\frac{\pi}{4}-\frac{\pi}{2}\right)=3\cos\!\left(2\pi t-\frac{\pi}{4}\right)\ \text{cm}. \] So \(\displaystyle A=3\ \text{cm}\), \(\displaystyle \omega=2\pi\ \text{rad s}^{-1}\), \(\displaystyle \phi=-\pi/4\ \text{rad}=-45^\circ\).Reference circle: radius \(\displaystyle 3\ \text{cm}\), anticlockwise at \(\displaystyle 2\pi\ \text{rad s}^{-1}\) (one full turn per second, since \(\displaystyle T=2\pi/\omega=1\ \text{s}\)). Starting point \[P_0=\left(3\cos(-45^\circ),3\sin(-45^\circ)\right)=\left(\frac{3\sqrt2}{2},-\frac{3\sqrt2}{2}\right)\ \text{cm}\approx(2.12,-2.12)\ \text{cm}, \] in the fourth quadrant, \(\displaystyle 45^\circ\) below the positive x-axis.(d) \(\displaystyle x=2\cos(\pi t)\)This is already exactly the standard form \(\displaystyle A\cos(\omega t+\phi)\) with \(\displaystyle \phi=0\): \(\displaystyle A=2\ \text{cm}\), \(\displaystyle \omega=\pi\ \text{rad s}^{-1}\), \(\displaystyle \phi=0\).Reference circle: radius \(\displaystyle 2\ \text{cm}\), anticlockwise at \(\displaystyle \pi\ \text{rad s}^{-1}\) (period \(\displaystyle T=2\pi/\pi=2\ \text{s}\)). Starting point \(\displaystyle P_0=(2,0)\ \text{cm}\) — the particle begins exactly on the positive x-axis.In every case, once \(\displaystyle P\) is placed on its circle it simply turns anticlockwise at the stated \(\displaystyle \omega\); dropping a perpendicular from \(\displaystyle P\) to the x-axis at any later time reproduces the given \(\displaystyle x(t)\).**Answer: (a) radius $\displaystyle 2$ cm, \(\displaystyle \omega=3\ \text{rad s}^{-1}\), starts at \(\displaystyle 150^\circ\) anticlockwise from the +x-axis (point \(\displaystyle (-1.73,1.00)\) cm). (b) radius $\displaystyle 1$ cm, \(\displaystyle \omega=1\ \text{rad s}^{-1}\), starts at \(\displaystyle -30^\circ\) (point \(\displaystyle (0.866,-0.500)\) cm). (c) radius $\displaystyle 3$ cm, \(\displaystyle \omega=2\pi\ \text{rad s}^{-1}\), starts at \(\displaystyle -45^\circ\) (point \(\displaystyle (2.12,-2.12)\) cm). (d) radius $\displaystyle 2$ cm, \(\displaystyle \omega=\pi\ \text{rad s}^{-1}\), starts at \(\displaystyle 0^\circ\) (point \(\displaystyle (2,0)\) cm). In every case the reference point rotates anticlockwise, and \(\displaystyle x(t)\) is its projection on the x-axis.

Splitting a spring at its midpoint doesn't change how hard each half pulls — it just relabels the anchor. That symmetry is the whole trick behind this problem.Case (a): one end clamped, force \(\displaystyle F\) on the free massA spring obeying Hooke's law stretches so that its tension equals the applied force: \[F = kx \] where \(\displaystyle k\) is the force constant of the spring and \(\displaystyle x\) is its extension. The clamped end does not move, so the entire displacement of the mass shows up as extension of the spring. Solving, \[x_a = \frac{F}{k}. \]When the mass is released, the spring pulls it back toward the natural length with restoring force \(\displaystyle -kx\) (Hooke's law again, now with the minus sign because the force opposes the displacement). Newton's second law gives \[m\frac{d^2x}{dt^2} = -kx \quad\Rightarrow\quad \omega_a = \sqrt{\frac{k}{m}}, \] which is the standard equation of simple harmonic motion. The period is \[T_a = \frac{2\pi}{\omega_a} = 2\pi\sqrt{\frac{m}{k}}. \]Case (b): both ends free, a mass \(\displaystyle m\) at each end, equal force \(\displaystyle F\) pulling each end outwardThe two applied forces are equal and opposite, and the two masses are identical, so the whole arrangement is symmetric about the spring's midpoint. Nothing distinguishes the two halves from each other, so that midpoint cannot move — by symmetry it behaves exactly like the rigid wall of case (a).Cut the spring there. A spring of force constant \(\displaystyle k\) and natural length \(\displaystyle L\), split into two equal pieces of length \(\displaystyle L/2\), becomes two springs of force constant \(\displaystyle 2k\) each — a shorter length stretches less for the same force, and since the extension per unit force is proportional to length, halving the length doubles the force constant. Each half is now exactly the case-(a) situation — one end anchored (at the now-motionless midpoint), the other pulled by force \(\displaystyle F\) — except the constant is \(\displaystyle 2k\) instead of \(\displaystyle k\). So each half stretches by \[x_{\text{half}} = \frac{F}{2k}. \] This is the step that trips people up — the quantity to compare with case (a) is the extension of the whole spring, which is the sum of both halves: \[x_b = 2 \times \frac{F}{2k} = \frac{F}{k}. \] So the maximum extension is the same in both cases, \(\displaystyle x_a = x_b = F/k\), even though case (b) has forces pulling at both ends instead of one. The doubled number of forces is exactly cancelled by the doubled stiffness of each half-spring — it is not simply "twice the stretch."When both masses are released, no external force acts on the system and it started at rest, so the centre of mass stays fixed for all time — it takes over the role the wall played in case (a). Each mass therefore oscillates against a half-spring of constant \(\displaystyle 2k\) anchored at that fixed centre: \[m\frac{d^2x}{dt^2} = -(2k)x \quad\Rightarrow\quad \omega_b = \sqrt{\frac{2k}{m}}. \] (The same result follows from treating the pair as a two-body oscillator with reduced mass \(\displaystyle \mu = \dfrac{m\cdot m}{m+m} = \dfrac{m}{2}\), giving \(\displaystyle \omega_b=\sqrt{k/\mu}=\sqrt{2k/m}\) — the two routes agree.) The period is \[T_b = \frac{2\pi}{\omega_b} = 2\pi\sqrt{\frac{m}{2k}} = \frac{T_a}{\sqrt{2}}. \]So the two-mass system swings back and forth faster than the single mass on a wall-mounted spring — even though a given end-force stretches it by exactly the same amount — because each mass is effectively anchored to a spring twice as stiff.Answer: Maximum extension is the same in both cases, \(\displaystyle x = F/k\). Period of oscillation: case (a) \(\displaystyle T_a = 2\pi\sqrt{m/k}\); case (b) \(\displaystyle T_b = 2\pi\sqrt{m/2k} = T_a/\sqrt{2}\).

In SHM, the speed is greatest at the centre of the oscillation, where \(\displaystyle v_{max} = A\omega \) — so the amplitude and the angular frequency are the only two numbers this problem needs.Step $\displaystyle 1$: Get the amplitude from the stroke.The stroke of a piston is the total distance it travels from one end of its motion to the other — that is twice the amplitude, not the amplitude itself. This is the detail people miss: reading "stroke = $\displaystyle 1.0$ m" and plugging it straight in as \(\displaystyle A\) doubles the final answer.\[\text{stroke} = 2A \implies A = \frac{1.0 \text{ m}}{2} = 0.50 \text{ m} \]Step $\displaystyle 2$: Convert the angular frequency to SI units.\(\displaystyle \omega\) is given as \(\displaystyle 200\) rad/min, but speed in m/s needs \(\displaystyle \omega\) in rad/s:\[\omega = \frac{200 \text{ rad}}{1 \text{ min}} \times \frac{1 \text{ min}}{60 \text{ s}} = \frac{200}{60} \text{ rad/s} = 3.33 \text{ rad/s} \]Step $\displaystyle 3$: Apply the maximum-speed formula for SHM.For displacement \(\displaystyle x(t) = A\sin(\omega t)\), the velocity is \(\displaystyle v(t) = A\omega\cos(\omega t)\), which is largest in magnitude when \(\displaystyle \cos(\omega t) = \pm 1\), giving\[v_{max} = A\omega \]Substituting the values from Steps $\displaystyle 1$ and $\displaystyle 2$:\[v_{max} = (0.50 \text{ m})\left(3.33 \text{ rad/s}\right) = 1.67 \text{ m/s} \](Working directly in the original units gives the same result without rounding early: \(\displaystyle v_{max} = A\omega = 0.50 \text{ m} \times 200 \text{ rad/min} = 100 \text{ m/min} = \dfrac{100}{60} \text{ m/s} = 1.67 \text{ m/s}\).)Rounding once, at the end, to three significant figures (matching the precision of the given angular frequency):Answer: The maximum speed of the piston is \(\displaystyle v_{max} \approx 1.67\ \text{m/s}\).

A pendulum's period depends on \(\displaystyle g\), so the same clock runs slower where gravity is weaker — you need the ratio of periods, not a fresh calculation from scratch.For a simple pendulum of length \(\displaystyle L\), the period is \[T = 2\pi\sqrt{\frac{L}{g}} \] where \(\displaystyle L\) is the length of the pendulum and \(\displaystyle g\) is the local acceleration due to gravity. The same pendulum (same \(\displaystyle L\)) is moved from Earth to the Moon, so \(\displaystyle L\) stays fixed while \(\displaystyle g\) changes — that fixed \(\displaystyle L\) is what lets you form a ratio and cancel it out.Write the period on Earth and on the Moon: \[T_E = 2\pi\sqrt{\frac{L}{g_E}}, \qquad T_M = 2\pi\sqrt{\frac{L}{g_M}} \]Dividing the two equations cancels \(\displaystyle 2\pi\) and \(\displaystyle L\): \[\frac{T_M}{T_E} = \sqrt{\frac{g_E}{g_M}} \]This is the key relation: period scales as the inverse square root of \(\displaystyle g\) — weaker gravity (smaller \(\displaystyle g_M\)) gives a longer period, which matches the everyday sense that a pendulum swings more lazily on the Moon.Substitute the given values \(\displaystyle T_E = 3.5\ \text{s}\), \(\displaystyle g_E = 9.8\ \text{m s}^{-2}\), \(\displaystyle g_M = 1.7\ \text{m s}^{-2}\): \[T_M = T_E\sqrt{\frac{g_E}{g_M}} = 3.5\ \text{s} \times \sqrt{\frac{9.8\ \text{m s}^{-2}}{1.7\ \text{m s}^{-2}}} \]Work out the ratio inside the root first: \[\frac{9.8}{1.7} = 5.76 \]Take the square root: \[\sqrt{5.76} = 2.4 \]Multiply: \[T_M = 3.5\ \text{s} \times 2.4 = 8.4\ \text{s} \]The input data (\(\displaystyle 3.5\), \(\displaystyle 9.8\), \(\displaystyle 1.7\)) each carry two significant figures, so the answer is rounded to two significant figures: \(\displaystyle 8.4\ \text{s}\), not the longer unrounded string of decimals the intermediate arithmetic produces.Note the direction of the effect: since the Moon's gravity is about one-sixth of Earth's, the pendulum's period on the Moon is longer, not shorter — a common slip is to divide the ratio the wrong way and predict a faster swing.Answer: The time period on the Moon is \(\displaystyle T_M \approx 8.4\ \text{s}\) (compared to \(\displaystyle 3.5\ \text{s}\) on Earth).

A pendulum swinging inside a turning car does not feel gravity alone — it feels gravity combined with the car's own acceleration.The car goes around a circular track of radius \(\displaystyle R\) at a constant speed \(\displaystyle v\), so it has a centripetal acceleration\[a = \frac{v^2}{R} \]directed horizontally, toward the centre of the track. Working in the reference frame of the car (a non-inertial frame, since it is turning), the bob of mass \(\displaystyle M\) appears to feel a pseudo-force\[F_{\text{pseudo}} = Ma = \frac{Mv^2}{R} \]pointing radially outward (away from the centre of the track) — this is the "centrifugal" push a passenger feels on a bend. At the same time the bob feels its real weight \(\displaystyle Mg\), acting vertically downward, where \(\displaystyle g\) is the acceleration due to gravity.These two forces add as vectors, not as numbers, because one is horizontal and the other vertical. The resultant force on the bob, in the car's frame, has magnitude\[F_{\text{eff}} = \sqrt{(Mg)^2 + \left(\frac{Mv^2}{R}\right)^2} = M\sqrt{g^2 + \frac{v^4}{R^2}} \]This resultant plays the role of an effective weight, so the bob behaves as though it hangs in an effective gravitational field of magnitude\[g_{\text{eff}} = \sqrt{g^2 + \frac{v^4}{R^2}} \]directed along the line of this resultant force — tilted away from the true vertical, toward the outside of the curve, by an angle \(\displaystyle \theta = \tan^{-1}\!\left(\dfrac{v^2}{Rg}\right)\). The pendulum's new equilibrium position (the position it hangs at rest, in the car's frame) lies along this tilted direction, not along the true vertical.The question specifies that the bob is displaced slightly from this equilibrium position, in the radial-vertical plane, and released — this is exactly the small-oscillation setup of a simple pendulum, except that the restoring field is \(\displaystyle g_{\text{eff}}\) instead of \(\displaystyle g\). For a simple pendulum of length \(\displaystyle l\) swinging through a small angle in a field of effective acceleration \(\displaystyle g_{\text{eff}}\), the restoring torque is linear in the displacement, giving simple harmonic motion with period\[T = 2\pi\sqrt{\frac{l}{g_{\text{eff}}}} \]exactly as for an ordinary pendulum, with \(\displaystyle g\) replaced by \(\displaystyle g_{\text{eff}}\). Substituting the expression found above,\[T = 2\pi\sqrt{\dfrac{l}{\sqrt{g^2 + \dfrac{v^4}{R^2}}}} \]Notice that the mass \(\displaystyle M\) has cancelled out completely — just as with an ordinary pendulum on solid ground, the period does not depend on the mass of the bob, only on the length \(\displaystyle l\) and the effective field \(\displaystyle g_{\text{eff}}\) it swings in. Also check the limiting case: if the car is going straight or is at rest, \(\displaystyle v = 0\), and the formula correctly collapses to the familiar \(\displaystyle T = 2\pi\sqrt{l/g}\).A common mistake here is to simply add \(\displaystyle g\) and \(\displaystyle v^2/R\) as plain numbers because both have units of acceleration — but they act in perpendicular directions, so they must be combined by the Pythagorean rule (as components of a vector sum), not added directly.Answer: \(\displaystyle T = 2\pi\sqrt{\dfrac{l}{\sqrt{g^2 + v^4/R^2}}}\), where the effective acceleration due to gravity felt by the bob is \(\displaystyle g_{\text{eff}} = \sqrt{g^2 + v^4/R^2}\), the vector combination of the true gravity \(\displaystyle g\) and the car's centripetal acceleration \(\displaystyle v^2/R\); the period is independent of the bob's mass \(\displaystyle M\).

A restoring force proportional to displacement is Newton's second law is SHM — the whole problem is finding that proportionality constant.Step $\displaystyle 1$: Equilibrium — Archimedes' principle.Let the cork (base area \(\displaystyle A\), height \(\displaystyle h\), density \(\displaystyle \rho\)) float with a length \(\displaystyle x_0\) submerged in the liquid (density \(\displaystyle \rho_l\)). Two forces act on it: its weight, pulling down, and the buoyant force — equal to the weight of liquid displaced — pushing up.Weight of cork: \[W = \rho A h g \]Buoyant force at equilibrium (volume submerged \(\displaystyle = A x_0\)): \[F_b = \rho_l A x_0 g \]Floating means these balance: \[\rho A h g = \rho_l A x_0 g \quad\Rightarrow\quad x_0 = \frac{\rho h}{\rho_l} \]This equilibrium depth itself is not needed in the final answer — it drops out — but writing it down is what lets the next step isolate the extra force cleanly.Step $\displaystyle 2$: Push the cork down by a small amount \(\displaystyle y\).Now the submerged length is \(\displaystyle x_0 + y\), so the buoyant force becomes \[F_b' = \rho_l A (x_0 + y) g = \rho_l A x_0 g + \rho_l A g\, y \]The first term, \(\displaystyle \rho_l A x_0 g\), is exactly the weight \(\displaystyle W\) — it was already balancing the cork. What's left over is a net upward force caused purely by the extra depth \(\displaystyle y\): \[F_{\text{net}} = W - F_b' = -\rho_l A g\, y \]The minus sign matters: when \(\displaystyle y\) is positive (cork pushed down), the net force points up, back toward equilibrium. This is the restoring force, and it is directly proportional to the displacement \(\displaystyle y\) — the defining signature of simple harmonic motion. (If the cork is instead pulled up slightly, the same expression with \(\displaystyle y<0\) correctly gives a net downward, restoring force.)Step $\displaystyle 3$: Apply Newton's second law.The oscillating mass is the cork itself, \(\displaystyle m = \rho A h\) (its density times its own volume — not the liquid's density; that mistake would give the wrong frequency). Newton's second law gives\[m\frac{d^2y}{dt^2} = -\rho_l A g\, y \] \[\rho A h\,\frac{d^2y}{dt^2} = -\rho_l A g\, y \]The cross-sectional area \(\displaystyle A\) cancels from both sides (it doesn't affect the frequency — a fatter cork of the same height oscillates at the same rate):\[\frac{d^2y}{dt^2} = -\left(\frac{\rho_l g}{\rho h}\right) y \]Step $\displaystyle 4$: Identify SHM and read off the period.This has the standard SHM form \(\displaystyle \dfrac{d^2y}{dt^2} = -\omega^2 y\), with\[\omega^2 = \frac{\rho_l g}{\rho h} \]Since the acceleration is a negative constant times the displacement, the motion is indeed simple harmonic, with angular frequency \(\displaystyle \omega = \sqrt{\dfrac{\rho_l g}{\rho h}}\). The period follows from \(\displaystyle T = \dfrac{2\pi}{\omega}\):\[T = 2\pi \sqrt{\frac{\rho h}{\rho_l g}} \]Here \(\displaystyle \rho\) is the density of the cork, \(\displaystyle \rho_l\) is the density of the liquid, \(\displaystyle h\) is the height of the cork, and \(\displaystyle g\) is the acceleration due to gravity. Every quantity under the root has units of density/density × length/(length/time²) = time², so the square root is a time, as a period must be — a useful check that no algebra step went wrong.Answer: The cork executes SHM with period \(\displaystyle T = 2\pi\sqrt{\dfrac{\rho h}{\rho_l g}}\), where \(\displaystyle \rho\) is the cork's density, \(\displaystyle \rho_l\) the liquid's density, \(\displaystyle h\) the cork's height, and \(\displaystyle g\) the acceleration due to gravity.

A restoring force that is proportional to displacement, and always directed back toward equilibrium, is the signature of SHM — the mercury's own weight supplies exactly that force here.Setting up the systemLet the mercury fill the U-tube to a total length \(\displaystyle L \) (measured along the tube, both limbs added together), let \(\displaystyle A \) be the (uniform) cross-sectional area of the bore, and let \(\displaystyle \rho \) be the density of mercury.With the suction pump removed, atmospheric pressure acts equally on both open ends, so the equilibrium configuration has the mercury standing at the same level in both limbs.Displacing the mercuryNow suppose the mercury is disturbed so that the level in one limb rises by \(\displaystyle x \) above the equilibrium mark. Mercury is incompressible and the tube has a single uniform bore, so the same volume that rose on one side must have left the other side — the level in the other limb falls by the same \(\displaystyle x \). The two limbs are then no longer level: the height difference between them is\[h = 2x \]This is the step people skip: the displacement of the mercury surface from its rest level is \(\displaystyle x \), but the height difference driving the flow is \(\displaystyle 2x \), because both arms move.The restoring forceThe extra column of mercury of height \(\displaystyle 2x \) sitting unbalanced on one side is what pushes the liquid back. Its weight is\[F = -(\text{weight of the unbalanced column}) = -(A \cdot 2x)\rho g \]\[F = -2A\rho g\,x \]The minus sign is there on physical grounds: this extra column always pushes the mercury back toward the level position, i.e. the force opposes the displacement \(\displaystyle x \).Newton's second law for the whole liquid columnThe entire mass of mercury moving together (it is one connected, incompressible column filling length \(\displaystyle L \) of the tube) is\[m = A L \rho \]Applying \(\displaystyle F = ma \):\[AL\rho\,\frac{d^2x}{dt^2} = -2A\rho g\,x \]The cross-section \(\displaystyle A \) and density \(\displaystyle \rho \) cancel from both sides — the motion does not depend on how wide the tube is or what liquid it is filled with, only on \(\displaystyle L \):\[\frac{d^2x}{dt^2} = -\left(\frac{2g}{L}\right)x \]Recognizing SHMCompare this with the defining equation of simple harmonic motion, \(\displaystyle \dfrac{d^2x}{dt^2} = -\omega^2 x \). The mercury's equation has exactly this form, with\[\omega^2 = \frac{2g}{L} \qquad\Rightarrow\qquad \omega = \sqrt{\frac{2g}{L}} \]Because the acceleration is proportional to the displacement \(\displaystyle x \) and directed opposite to it at every instant, the mercury column executes simple harmonic motion about its equilibrium (level) position, with angular frequency \(\displaystyle \omega = \sqrt{2g/L} \) and period\[T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L}{2g}} \]where \(\displaystyle L \) is the total length of the mercury column in the tube (not the height of mercury in one arm alone — using just one arm's length is the usual error here, and it silently drops the factor of $\displaystyle 2$ as well).**Answer: Displacing the mercury by \(\displaystyle x\) in one limb creates a height difference of \(\displaystyle 2x\) between the two columns, giving a restoring force \(\displaystyle F = -2A\rho g\,x\) on the mercury of total length \(\displaystyle L\). Newton's second law then gives \(\displaystyle \dfrac{d^2x}{dt^2} = -\left(\dfrac{2g}{L}\right)x\), which is the SHM equation with \(\displaystyle \omega = \sqrt{2g/L}\) — so the mercury column indeed executes simple harmonic motion, with period \(\displaystyle T = 2\pi\sqrt{L/2g}\).