Q.The length of a second's pendulum on the surface of Earth is 1m. What will be the length of a second's pendulum on the moon?

Key idea. "$\tfrac18$ of an oscillation" is a fixed fraction of the full cycle, so the time each particle needs is $\dfrac{T}{8}$, independent of amplitude. The amplitudes ($10$ cm, $20$ cm) do not affect the time for a given fraction of the pe …

The key here is to recognise that a floating object, when displaced vertically, experiences a net restoring force proportional to the displacement — exactly like a spring. This is because the extra volume submerged (or unsubmerged) changes the buoyant force linearly with depth.

Let’s work through it.

1. Set up the physics of the oscillation

   The cork is floating in equilibrium. When you push it down by a small distance $x$, an extra volume $A x$ of water is displaced, where $A$ is the cross-sectional area of the cube. The additional buoyant force upward is $\rho_{\text{water}} \cdot (A x) \cdot g$. This force opposes the displacement, so the net restoring force is $F = - (\rho A g) x$.

   This is exactly Hooke’s law, $F = -k x$, with effective spring constant $k = \rho A g$.

2. Relate time period to mass

   For SHM, the time period is $T = 2\pi \sqrt{\frac{m}{k}}$. Substituting $k$:

3. Plug in the numbers Side of cube = 10 cm = 0.1 m, so $A = (0.1)^2 = 0.01 \, \text{m}^2$. Density of water $\rho = 1000 \, \text{kg/m}^3$, $g = 10 \, \text{m/s}^2$, and $T = 0.5 \, \text{s}$. Square both sides:

Substitute: …

The problem gives $x = 18 \sin \left( 2\pi t + \frac{\pi}{2} \right)$ cm. The first thing to notice is that $\sin(\theta + \pi/2) = \cos \theta$. So the displacement is actually $x = 18 \cos(2\pi t)$ cm. That means at $t = 0$, $x = 18$ cm — the particle starts at the positive extreme position.

In simple harmonic motion, velocity is maximum when the particle passes through the mean position ($x = 0$). So we need the smallest positive time after $t = 0$ when $x = 0$.

1. Find the period. The angular frequency $\omega = 2\pi$ rad/s. The time period $T = \frac{2\pi}{\omega} = \frac{2\pi}{2\pi} = 1$ s.

2. Locate the first zero. Starting from the extreme ($x = 18$ at $t = 0$), the particle moves toward the mean. It reaches $x = 0$ for the first time after one-quarter of a period. Why? Because in SHM, starting from an extreme, the motion to the mean takes $T/4$.

3. Compute the time. $t = \frac{T}{4} = \frac{1}{4} = 0.25$ s. …

The key idea is that a vehicle topples when the net torque about the pivot point (the outer wheel) becomes unbalanced. As the truck goes around a curve, the centrifugal force (apparent outward force in the truck’s frame) tries to tip it outward, while gravity tries to keep it upright. The maximum safe speed is when these two torques are exactly equal — any faster and the truck will start to lift its inner wheels.

1. Identify the pivot and the forces

   When the truck is about to topple, the inner wheels lose contact with the ground. The entire weight acts through the center of gravity (CG), and the pivot is the outer wheel. The relevant forces are:

   • Weight $mg$ acting downward at the CG.
   • Centrifugal force $\frac{mv^2}{r}$ acting horizontally outward at the CG (in the rotating frame).

2. Set up the torque balance about the outer wheel

   Take torques about the point where the outer wheel touches the ground.

   • The weight creates a torque that tries to restore the truck to upright: its lever arm is half the wheel track, $\frac{1.5}{2} = 0.75\ \mathrm{m}$.

• The centrifugal force tries to topple the truck: its lever arm is the height of the CG above ground, $2\ \mathrm{m}$.

3. Equate the torques at the tipping point At the maximum speed without toppling, these torques are equal: $mg \times 0.75 = \frac{mv^2}{r} \times 2$ …

The key idea is to recall the standard result for a circular loop’s moment of inertia about a diameter, then express the radius in terms of the given mass and linear density.

Why this works:

For a thin uniform wire bent into a circle, every bit of mass lies at the same distance $R$ from the center. The moment of inertia about a diameter is a classic result: because the mass is distributed symmetrically, the perpendicular-axis theorem gives $I_{\text{diameter}} = \frac12 I_{\text{axis through center perpendicular to plane}}$, and $I_{\text{perpendicular}} = mR^2$. So $I_{\text{diameter}} = \frac12 mR^2$. The only twist is that the problem gives linear mass density $\rho$, not the radius directly — so we must relate $R$ to $m$ and $\rho$.

Step-by-step:

1. Relate mass, length, and radius.

   The wire is bent into a circle, so its total length is the circumference $2\pi R$. Linear mass density $\rho$ means mass per unit length: $\rho = \frac{m}{\text{length}} = \frac{m}{2\pi R}$.

   Hence, $R = \frac{m}{2\pi \rho}$.

2. Use the known moment of inertia for a loop about a diameter.

   For a thin circular loop of mass $m$ and radius $R$, the moment of inertia about any diameter is $I = \frac12 m R^2$. (This follows from the perpendicular-axis theorem: $I_z = mR^2$, and by symmetry $I_x = I_y$, so $I_x + I_y = I_z$ gives $2I_x = mR^2$, hence $I_x = \frac12 mR^2$.)

3. Substitute $R$ from step 1 into $I$. …

1. Speed at the highest point. Zero tension means gravity alone supplies the centripetal force:

2. Maximum kinetic energy (at the lowest point). Using energy conservation over the diameter $2R$:

3. Where KE is half of maximum. Let $\theta$ be the angle of the string from the upward vertical (measured at the top). Height of that point above the lowest point is $R(1+\cos\theta)$. Energy conservation from the bottom: $KE(\theta) = KE_{\max} - mgR(1+\cos\theta).$ …

Concept & Intuition

A pendulum in a bus moving uniformly on a horizontal circular road is not an inertial frame — the bus is accelerating toward the center of the circle. From the bus’s perspective, the bob feels a centrifugal pseudo-force outward, horizontal and of magnitude $mV^2/R$. This combines with gravity to give an effective weight. The wire’s tension equals the magnitude of this effective weight, and the elongation is given by $\Delta L = \frac{T L}{A Y}$ (Hooke’s law for a wire under tension). The key is to find the net force on the bob correctly.

Step-by-step reasoning

1. Identify the forces in the bus’s frame

   In the non-inertial frame of the bus, the bob is at rest relative to the bus. The real forces are gravity ($mg$ downward) and tension ($T$ along the wire). The pseudo-force is the centrifugal force: $F_c = \frac{mV^2}{R}$, directed radially outward from the center of the circular path. This force is horizontal and perpendicular to the vertical.

2. Resultant effective force on the bob

   The bob is in equilibrium in the bus frame, so tension must balance the vector sum of gravity and centrifugal force. These two forces are perpendicular:

   • Vertical: $mg$
   • Horizontal: $\frac{mV^2}{R}$ The magnitude of the resultant (effective weight) is

3. Tension in the wire equals this effective force Since the bob is stationary in the bus frame, the tension $T$ must exactly oppose $F_{\text{eff}}$, so

4. Apply Hooke’s law for elongation For a wire of length $L$, cross-sectional area $A$, and Young’s modulus $Y$, the elongation $\Delta L$ is $\Delta L = \frac{T L}{A Y}.$ …

The concept first

Two separate ideas meet here, and it pays to keep them apart.

Idea 1 — the swing is just energy conversion. A pendulum bob released from an angle $\theta$ has no engine; gravity converts its lost height into speed, and on the way up the reverse happens. Nothing else matters — not the mass, not the path — only the vertical drop.

Idea 2 — the elastic equal-mass collision is the cleanest result in all of mechanics. Solve momentum + kinetic-energy conservation for a head-on elastic collision of masses $m_1,m_2$ and you get

Put $m_1=m_2=m$: the first bracket is zero and the second is one. The incoming bob stops; the struck bob inherits the entire velocity. This is what you see in a Newton's-cradle.

Step-by-step

1. Height P falls. With the string of length $L$ making $60^\circ$ with the vertical, the bob starts a vertical distance $L\cos 60^\circ$ below the support, and ends (at the lowest point) a distance $L$ below it. So the drop is

2. Speed of P just before impact. The string tension does no work (it is always perpendicular to the motion), so mechanical energy is conserved:

3. The collision. Masses are equal and the collision is elastic, so from the boxed result above

Q is set moving horizontally with speed $\sqrt{gL}$; P is left hanging at rest. …

The key idea is that when the car moves on a circular track, the bob experiences a centrifugal acceleration outward (in the rotating frame of the car). For small oscillations in the radial direction, the pendulum’s equilibrium is tilted, and the restoring force depends on the net effective gravity — the vector sum of $g$ downward and the centrifugal acceleration $v^2/r$ horizontally outward.

The problem gives the time period as $T = 2\pi / \alpha^{1/4}$. For a simple pendulum, the standard formula is $T = 2\pi \sqrt{L/g_{\text{eff}}}$. So we must find $g_{\text{eff}}$ and match the form.

1. Find the centrifugal acceleration. The car’s speed is $v = 10\ \text{m/s}$ and the track radius is $R = 100\ \text{m}$. The centrifugal acceleration (in the car’s frame) is

This acts radially outward, perpendicular to gravity $g = 10\ \text{m/s}^2$ downward.

2. Determine the effective gravity. The bob’s equilibrium is along the direction of the vector sum of $g$ and $a_c$. The magnitude of the effective gravity is

3. Write the time period for small radial oscillations. For a pendulum of length $L = 1\ \text{m}$ oscillating about this tilted equilibrium, the small-angle period is …

We start with the physics: In simple harmonic motion, total energy $E$ is constant, split between kinetic energy $K$ and potential energy $U$. When $K = 0.75E$, then $U = 0.25E$. Since potential energy in SHM is proportional to the square of displacement ($U = \frac12 k x^2$), and total energy is $E = \frac12 k A^2$ (where $A$ is amplitude), we have:

So the question reduces to: starting from the mean position ($x=0$) at $t=0$, how long does it take to reach $x = A/2$ for the first time?

1. Write the equation of motion. For SHM starting from the mean position moving in the positive direction, we use $x = A \sin(\omega t)$. Here $\omega = \frac{2\pi}{T}$, and $T = 4$ s, so $\omega = \frac{2\pi}{4} = \frac{\pi}{2}$ rad/s.

2. Set displacement to half the amplitude. We want $x = \frac{A}{2}$, so:

   $\frac{A}{2} = A \sin\left(\frac{\pi}{2} t\right) \quad \Rightarrow \quad \sin\left(\frac{\pi}{2} t\right) = \frac12.$ …

The key idea: average acceleration is defined as $\vec{a}_{\text{avg}} = \frac{\Delta \vec{v}}{\Delta t}$. In a uniform magnetic field, the particle moves in a circular arc at constant speed $V$, so the magnitude of the velocity change over a $90^\circ$ turn is easy to find geometrically, and the time is just a fraction of the full cyclotron period.

Why this works:

The magnetic force is always perpendicular to velocity, so speed is constant. The path is a circle of radius $R = \frac{mV}{qB}$. Over a $90^\circ$ arc, the velocity vector rotates by $90^\circ$, so the magnitude of the change in velocity is $|\Delta \vec{v}| = \sqrt{V^2 + V^2} = \sqrt{2}\,V$. The time to cover a quarter-circle is one-quarter of the period $T = \frac{2\pi m}{qB}$, so $\Delta t = \frac{T}{4} = \frac{\pi m}{2qB}$. Dividing gives the average acceleration.

1. Find the change in velocity $\Delta \vec{v}$. At point 1, velocity is, say, $\vec{v}_1 = V\,\hat{i}$. After a $90^\circ$ turn to point 2, velocity is $\vec{v}_2 = V\,\hat{j}$ (or some perpendicular direction). The vector difference is $\vec{v}_2 - \vec{v}_1 = V\hat{j} - V\hat{i}$, whose magnitude is

2. Find the time $\Delta t$ for the $90^\circ$ turn. In a uniform magnetic field, the cyclotron period is $T = \frac{2\pi m}{qB}$. A $90^\circ$ turn is one-quarter of a full circle, so

3. Compute the average acceleration magnitude. $$$$

The key insight is that you cannot treat the rod as a point mass because the distance from the point mass to different parts of the rod varies. The rod is extended, so each tiny piece of it exerts a slightly different gravitational pull. To get the total force, we must add up (integrate) the contributions from all these pieces.

Why the simple formulas fail

• Option (A) $\dfrac{GMm}{r^2}$ would be correct only if the entire rod's mass were concentrated at its nearest end — but the rest of the rod is farther away, so the actual force is smaller.
• Option (B) $\dfrac{GMm}{(r+L)^2}$ would be correct if the mass were at the far end — but that underestimates the force because much of the rod is closer.
• Option (D) $\dfrac{GMm}{(r+L/2)^2}$ treats the rod as a point mass at its center. This is a common approximation, but it is exact only for a sphere or when $r \gg L$. For a rod at close range, the distribution matters.

So we must integrate.

Step-by-step derivation

1. Set up coordinates

   Place the rod along the $x$-axis from $x = 0$ to $x = L$. Let the point mass $m$ be on the same line at $x = -r$ (so the distance from $m$ to the near end of the rod is $r$).

   Then a small piece of the rod at position $x$ (where $0 \le x \le L$) has length $dx$ and mass $dM = \frac{M}{L} dx$ (linear mass density $\lambda = M/L$).

2. Distance from point mass to the piece

   The distance from $m$ (at $x = -r$) to the piece at $x$ is $x - (-r) = x + r$.

3. Gravitational force from one piece

   By Newton’s law of gravitation, the magnitude of the force between $m$ and this infinitesimal mass $dM$ is

4. Integrate over the whole rod The total force is the sum of all these contributions:

The constants $G$, $m$, $M$, and $L$ come out:

$F = \frac{G m M}{L} \int_{0}^{L} \frac{dx}{(x+r)^2}.$ …