Welcome to Lecture 106 of our comprehensive JEE Physics Master Series. In this lecture, we formulate the rotational mechanics of oscillating systems: the Simple Pendulum, the Physical (Compound) Pendulum, and the Torsional Oscillator. We examine the small-angle approximation, derive the center of oscillation and conjugate suspension points, analyze effective gravity in accelerating frames, and deliver 4 rigorous solved examples plus a 10-question practice worksheet (DPP) with complete solutions.
1. The Simple Pendulum & The Small-Angle Approximation
A simple pendulum consists of a point-mass bob of mass $m$ suspended by an inextensible, massless string of length $L$. When displaced by an angle $\theta$ from the vertical, the restoring torque about the pivot axis $O$ is:
$$\tau = -m g L \sin\theta$$
Rotational form of Newton’s second law ($\tau = I \alpha = I \frac{d^2\theta}{dt^2}$), with moment of inertia $I = m L^2$:
$$m L^2 \frac{d^2\theta}{dt^2} + m g L \sin\theta = 0 \implies \frac{d^2\theta}{dt^2} + \frac{g}{L}\sin\theta = 0$$
For small angular displacements ($\theta \ll 1\text{ rad}$, typically $\theta \le 4^\circ \approx 0.07\text{ rad}$), we take the Taylor approximation $\sin\theta \approx \theta$:
$$\frac{d^2\theta}{dt^2} + \left(\frac{g}{L}\right)\theta = 0 \implies \omega = \sqrt{\frac{g}{L}}, \quad T = 2\pi\sqrt{\frac{L}{g}}$$
Large-Amplitude Borda Correction: When angular amplitude $\theta_0$ is not negligible, the second-order Taylor expansion yields:
$$T \approx 2\pi\sqrt{\frac{L}{g}}\left(1 + \frac{1}{4}\sin^2\left(\frac{\theta_0}{2}\right)\right) \approx 2\pi\sqrt{\frac{L}{g}}\left(1 + \frac{\theta_0^2}{16}\right)$$
Thus, the time period increases slightly as amplitude increases.
2. Simple Pendulum in Non-Inertial Reference Frames
In a reference frame accelerating with linear acceleration $\vec{a}_0$, the effective gravitational field vector is:
$$\vec{g}_{\text{eff}} = \vec{g} – \vec{a}_0 \implies g_{\text{eff}} = |\vec{g} – \vec{a}_0|$$
The time period of small oscillations becomes:
$$T = 2\pi\sqrt{\frac{L}{g_{\text{eff}}}}$$
Common cases:
- Elevator accelerating upward ($a_0$): $g_{\text{eff}} = g + a_0 \implies T = 2\pi\sqrt{\frac{L}{g + a_0}}$
- Elevator accelerating downward ($a_0 < g$): $g_{\text{eff}} = g – a_0 \implies T = 2\pi\sqrt{\frac{L}{g – a_0}}$
- Free-fall ($a_0 = g$): $g_{\text{eff}} = 0 \implies T \to \infty$ (pendulum does not oscillate)
- Car accelerating horizontally ($a_0$): $g_{\text{eff}} = \sqrt{g^2 + a_0^2}$, mean position tilted at angle $\beta = \tan^{-1}(a_0/g)$
3. The Physical (Compound) Pendulum
A physical pendulum is any rigid body of mass $m$ free to oscillate about a fixed horizontal axis passing through suspension point $S$. Let the center of mass $C$ be at a distance $\ell$ from $S$, and let $I_S$ be the moment of inertia about the pivot axis.
Restoring torque for angular deflection $\theta$:
$$\tau = -m g \ell \sin\theta \approx -m g \ell \theta$$
The equation of motion is:
$$I_S \frac{d^2\theta}{dt^2} + m g \ell \theta = 0 \implies \omega = \sqrt{\frac{m g \ell}{I_S}}, \quad T = 2\pi\sqrt{\frac{I_S}{m g \ell}}$$
Using the Parallel Axis Theorem, $I_S = I_{\text{cm}} + m \ell^2 = m k_g^2 + m \ell^2 = m(k_g^2 + \ell^2)$, where $k_g$ is the radius of gyration about the center-of-mass axis:
$$T = 2\pi\sqrt{\frac{m(k_g^2 + \ell^2)}{m g \ell}} = 2\pi\sqrt{\frac{k_g^2 + \ell^2}{g \ell}} = 2\pi\sqrt{\frac{L_{\text{eq}}}{g}}$$
where $L_{\text{eq}} = \ell + \frac{k_g^2}{\ell}$ is the equivalent length of a simple pendulum.
Center of Oscillation & Conjugate Points:
A point $O$ lying on the line joining $S$ and $C$ at distance $L_{\text{eq}}$ from $S$ is called the center of oscillation. If the rigid body is suspended from $O$, its new distance to the center of mass is $\ell’ = L_{\text{eq}} – \ell = \frac{k_g^2}{\ell}$. The time period about $O$ is identical to that about $S$ ($T_O = T_S$). Thus, $S$ and $O$ are mutually interchangeable conjugate points.
Condition for Minimum Time Period:
Minimizing $L_{\text{eq}}(\ell) = \ell + \frac{k_g^2}{\ell}$ with respect to $\ell$:
$$\frac{dL_{\text{eq}}}{d\ell} = 1 – \frac{k_g^2}{\ell^2} = 0 \implies \ell = k_g$$
$$T_{\text{min}} = 2\pi\sqrt{\frac{2 k_g}{g}}$$
4. The Torsional Oscillator
A torsional oscillator consists of a body of moment of inertia $I$ suspended by a vertical wire. When rotated by an angle $\theta$, the wire is twisted and exerts a restoring torque proportional to the angle of twist:
$$\tau = -C \theta$$
where $C$ is the torsional rigidity (torsional constant) of the wire ($\text{N}\cdot\text{m/rad}$). For a solid cylindrical wire of shear modulus $\eta$, radius $r$, and length $L$:
$$C = \frac{\pi \eta r^4}{2 L}$$
Governing equation of torsional SHM:
$$I \frac{d^2\theta}{dt^2} + C \theta = 0 \implies \omega = \sqrt{\frac{C}{I}}, \quad T = 2\pi\sqrt{\frac{I}{C}}$$
5. Solved Examples
Example 1 (Foundational – JEE Main):
A simple pendulum has a time period of $2.0\text{ s}$ inside a stationary elevator. Calculate its time period when the elevator accelerates upward at $a_0 = 2.0\text{ m/s}^2$. Take $g = 9.8\text{ m/s}^2$.
Solution:
When stationary, $T_0 = 2\pi\sqrt{\frac{L}{g}} = 2.0\text{ s}$.
When accelerating upward, effective gravity is $g_{\text{eff}} = g + a_0 = 9.8 + 2.0 = 11.8\text{ m/s}^2$.
$$T = 2\pi\sqrt{\frac{L}{g_{\text{eff}}}} = T_0 \sqrt{\frac{g}{g + a_0}} = 2.0 \times \sqrt{\frac{9.8}{11.8}} = 2.0 \times \sqrt{0.8305} \approx 2.0 \times 0.9113 = \mathbf{1.82\text{ s}}$$
Example 2 (Uniform Rod Physical Pendulum – JEE Main/Advanced):
A uniform thin rod of mass $M$ and length $L$ is pivoted at one of its ends and allowed to oscillate in a vertical plane with small amplitude. (a) Determine its time period $T$. (b) Find the position of the center of oscillation. (c) At what distance from the center of mass should the pivot be located to obtain the minimum time period?
Solution:
(a) Distance from pivot to center of mass is $\ell = L/2$.
Moment of inertia about end pivot: $I = \frac{1}{3} M L^2$.
$$T = 2\pi\sqrt{\frac{I}{M g \ell}} = 2\pi\sqrt{\frac{\frac{1}{3} M L^2}{M g (L/2)}} = \mathbf{2\pi\sqrt{\frac{2L}{3g}}}$$
(b) Comparing with $T = 2\pi\sqrt{\frac{L_{\text{eq}}}{g}}$, the equivalent length is $L_{\text{eq}} = \frac{2}{3}L$.
The center of oscillation lies along the rod at a distance $\mathbf{\frac{2}{3}L}$ from the top pivot point.
(c) For a uniform rod, radius of gyration about CM is $k_g = \sqrt{\frac{I_{\text{cm}}}{M}} = \sqrt{\frac{M L^2 / 12}{M}} = \frac{L}{\sqrt{12}} = \frac{L}{2\sqrt{3}}$.
The minimum time period occurs when $\ell = k_g = \mathbf{\frac{L}{2\sqrt{3}} \approx 0.289 L}$ from the center of mass.
Minimum period: $T_{\text{min}} = 2\pi\sqrt{\frac{2 k_g}{g}} = 2\pi\sqrt{\frac{2 L / (2\sqrt{3})}{g}} = 2\pi\sqrt{\frac{L}{\sqrt{3}g}}$.
Example 3 (Ring Pivoted on Knife Edge – JEE Advanced):
A uniform circular ring of mass $M$ and radius $R$ is hung from a sharp horizontal knife edge passing through its circumference, such that it can oscillate in: (a) its own vertical plane, and (b) a vertical plane perpendicular to the ring’s plane. Find the ratio of the time periods in these two oscillation modes.
Solution:
In both cases, the distance from pivot to center of mass is $\ell = R$.
(a) Oscillations in the plane of the ring:
Moment of inertia about CM perpendicular to ring plane is $I_{\text{cm}} = M R^2$.
By parallel axis theorem, pivot axis is also perpendicular to ring plane: $I_1 = I_{\text{cm}} + M R^2 = M R^2 + M R^2 = 2 M R^2$.
$$T_1 = 2\pi\sqrt{\frac{I_1}{M g \ell}} = 2\pi\sqrt{\frac{2 M R^2}{M g R}} = 2\pi\sqrt{\frac{2R}{g}}$$
(b) Oscillations perpendicular to the plane of the ring:
Here the pivot axis lies in the plane of the ring tangent to its circumference.
By perpendicular axis theorem, moment of inertia about diameter is $I_{\text{dia}} = \frac{1}{2} M R^2$.
By parallel axis theorem: $I_2 = I_{\text{dia}} + M R^2 = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2$.
$$T_2 = 2\pi\sqrt{\frac{I_2}{M g \ell}} = 2\pi\sqrt{\frac{\frac{3}{2} M R^2}{M g R}} = 2\pi\sqrt{\frac{3R}{2g}}$$
Ratio of time periods:
$$\frac{T_1}{T_2} = \frac{\sqrt{2R/g}}{\sqrt{3R/2g}} = \sqrt{\frac{2}{3/2}} = \sqrt{\frac{4}{3}} = \mathbf{\frac{2}{\sqrt{3}}}$$
Example 4 (Torsional Pendulum with Added Mass – JEE Advanced):
A uniform disc of mass $M$ and radius $R$ is suspended horizontally by a wire attached to its center. It executes torsional oscillations with time period $T_1 = 4.0\text{ s}$. Two small identical point masses, each of mass $m$, are placed on the disc at distance $r$ from the axis along a diameter, and the period increases to $T_2 = 5.0\text{ s}$. When both masses are moved to the outer rim of the disc ($r = R$), the period becomes $T_3 = 6.0\text{ s}$. Find the ratio of masses $m / M$.
Solution:
Initial moment of inertia: $I_0 = \frac{1}{2} M R^2$.
Initial period: $T_1 = 2\pi\sqrt{\frac{I_0}{C}} \implies T_1^2 \propto I_0 = \frac{1}{2} M R^2$.
When masses are at distance $R$ (at the rim):
$$I_{\text{rim}} = I_0 + 2 m R^2 = \frac{1}{2} M R^2 + 2 m R^2 = \left(\frac{M}{2} + 2m\right)R^2$$
Period ratio between state 3 and state 1:
$$\frac{T_3^2}{T_1^2} = \frac{I_{\text{rim}}}{I_0} = \frac{\left(\frac{M}{2} + 2m\right)R^2}{\frac{1}{2} M R^2} = 1 + \frac{2m}{M/2} = 1 + \frac{4m}{M}$$
Substitute $T_1 = 4.0\text{ s}$ and $T_3 = 6.0\text{ s}$:
$$\frac{6.0^2}{4.0^2} = \frac{36}{16} = \frac{9}{4} = 2.25$$
$$1 + \frac{4m}{M} = 2.25 \implies \frac{4m}{M} = 1.25 = \frac{5}{4} \implies \frac{m}{M} = \frac{5}{16} = \mathbf{0.3125}$$
6. 10-Question Practice Worksheet (DPP)
Q1 (Single Choice): A simple pendulum has time period $T$. If the length of the pendulum is increased by $21\%$, the percentage increase in its time period is:
(A) $10\%$
(B) $21\%$
(C) $42\%$
(D) $11\%$
Q2 (Single Choice): A simple pendulum suspended from the ceiling of a train has time period $T$ when the train is at rest. When the train accelerates horizontally with constant acceleration $a = g$, the new time period of the pendulum is:
(A) $T$
(B) $T/2$
(C) $T / 2^{1/4}$
(D) $2^{1/4} T$
Q3 (Single Choice): The length of a simple seconds pendulum ($T = 2\text{ s}$) on the surface of the Earth ($g = \pi^2\text{ m/s}^2$) is:
(A) $0.5\text{ m}$
(B) $1.0\text{ m}$
(C) $2.0\text{ m}$
(D) $4.0\text{ m}$
Q4 (Single Choice): A uniform circular disc of radius $R$ oscillates as a physical pendulum in a vertical plane about a horizontal axis through a point on its rim. The time period of small oscillations is:
(A) $2\pi\sqrt{\frac{R}{g}}$
(B) $2\pi\sqrt{\frac{2R}{3g}}$
(C) $2\pi\sqrt{\frac{3R}{2g}}$
(D) $2\pi\sqrt{\frac{R}{2g}}$
Q5 (Single Choice): For a physical pendulum of radius of gyration $k_g$ about its center of mass, the minimum time period of oscillation is achieved when the distance $\ell$ between the suspension point and the center of mass equals:
(A) $k_g / 2$
(B) $k_g$
(C) $2 k_g$
(D) $k_g^2$
Q6 (Single Choice): The torsional constant of a wire is $C$. If another wire of the same material has half the length and double the radius, its torsional constant is:
(A) $2C$
(B) $8C$
(C) $16C$
(D) $32C$
Q7 (One or More than One Correct): Which of the following statements regarding a simple pendulum are strictly TRUE?
(A) For very large amplitudes, the motion is periodic but not simple harmonic.
(B) In a satellite orbiting Earth in a circular orbit, the simple pendulum does not oscillate ($T \to \infty$).
(C) If the mass of the bob is doubled, the time period remains unchanged.
(D) If a hollow bob filled with water develops a leak at the bottom, the time period first increases and then returns to its initial value.
Q8 (One or More than One Correct): A physical pendulum consists of a rigid body oscillating about a horizontal axis. Then:
(A) There exist two collinear points on either side of the center of mass that have identical time periods.
(B) The center of oscillation and point of suspension are mutually conjugate.
(C) The equivalent length of a simple pendulum is always greater than or equal to $2k_g$.
(D) The time period is independent of the acceleration due to gravity.
Q9 (Numerical Value): A uniform meter stick ($L = 100\text{ cm}$) is pivoted at the $20\text{ cm}$ mark. Taking $g = 9.8\text{ m/s}^2$, the equivalent length of a simple pendulum $L_{\text{eq}}$ in $\text{cm}$ is ________.
Q10 (Numerical Value): A solid cylinder of mass $2\text{ kg}$ and radius $0.1\text{ m}$ is attached to a torsion wire of torsional rigidity $C = 0.05\text{ N}\cdot\text{m/rad}$ along its central cylindrical axis. The time period of torsional oscillation in seconds (rounded to two decimal places) is ________. (Take $\pi = 3.1416$)
7. Answer Key & Detailed Step-by-Step Solutions
Answer Key:
Q1: (A) | Q2: (C) | Q3: (B) | Q4: (C) | Q5: (B) | Q6: (D) | Q7: (A, B, C, D) | Q8: (A, B, C) | Q9: 57.78 (or $520/9$) | Q10: 2.81
Detailed Solutions
Solution 1:
$T = 2\pi\sqrt{L/g} \propto \sqrt{L}$.
If $L’ = 1.21 L$, then $T’ = \sqrt{1.21} T = 1.10 T$.
Percentage increase: $\frac{T’ – T}{T} \times 100\% = (1.10 – 1.00) \times 100\% = \mathbf{10\%}$. Hence, (A) is correct.
Solution 2:
Effective gravity in train: $g_{\text{eff}} = \sqrt{g^2 + a^2} = \sqrt{g^2 + g^2} = g\sqrt{2} = 2^{1/2} g$.
New period: $T’ = 2\pi\sqrt{\frac{L}{g_{\text{eff}}}} = 2\pi\sqrt{\frac{L}{2^{1/2} g}} = \frac{T}{2^{1/4}}$. Hence, (C) is correct.
Solution 3:
For seconds pendulum, $T = 2\text{ s}$.
$$T = 2\pi\sqrt{\frac{L}{g}} \implies 2 = 2\pi\sqrt{\frac{L}{\pi^2}} \implies 1 = \pi \frac{\sqrt{L}}{\pi} = \sqrt{L} \implies L = \mathbf{1.0\text{ m}}$$
Hence, (B) is correct.
Solution 4:
For a circular disc pivoted at the rim: $\ell = R$.
Moment of inertia about perpendicular rim axis: $I = I_{\text{cm}} + M R^2 = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2$.
$$T = 2\pi\sqrt{\frac{I}{M g \ell}} = 2\pi\sqrt{\frac{\frac{3}{2} M R^2}{M g R}} = 2\pi\sqrt{\frac{3R}{2g}}$$
Hence, (C) is correct.
Solution 5:
Equivalent length is $L_{\text{eq}} = \ell + \frac{k_g^2}{\ell}$.
Differentiating with respect to $\ell$: $\frac{dL_{\text{eq}}}{d\ell} = 1 – \frac{k_g^2}{\ell^2} = 0 \implies \ell = k_g$. Hence, (B) is correct.
Solution 6:
Torsional constant: $C = \frac{\pi \eta r^4}{2L} \propto \frac{r^4}{L}$.
With $r’ = 2r$ and $L’ = L/2$:
$$C’ \propto \frac{(2r)^4}{L/2} = \frac{16 r^4}{L/2} = 32 \frac{r^4}{L} = 32 C$$
Hence, (D) is correct.
Solution 7:
(A) True: Restoring torque is $-mgL\sin\theta$; without $\sin\theta \approx \theta$, the restoring force is non-linear.
(B) True: An orbiting satellite is in free-fall ($g_{\text{eff}} = 0 \implies T \to \infty$).
(C) True: $T = 2\pi\sqrt{L/g}$ is mass-independent.
(D) True: As water leaks, the center of mass descends, increasing effective length $L$ and period $T$. When fully drained, the center of mass returns to the center of the sphere, restoring the initial time period.
Hence, (A, B, C, D) are all correct.
Solution 8:
(A) True: For any period $T$, $L_{\text{eq}} = \ell + k_g^2/\ell \implies \ell^2 – L_{\text{eq}}\ell + k_g^2 = 0$, giving two positive roots $\ell_1$ and $\ell_2 = k_g^2/\ell_1$ on either side of the CM.
(B) True: Swapping suspension point $S$ and center of oscillation $O$ yields identical periods.
(C) True: By AM-GM inequality, $L_{\text{eq}} = \ell + \frac{k_g^2}{\ell} \ge 2\sqrt{\ell \cdot \frac{k_g^2}{\ell}} = 2k_g$.
(D) False: $T \propto 1/\sqrt{g}$, so it depends on gravity.
Hence, (A, B, C) are correct.
Solution 9:
Center of mass of a uniform $100\text{ cm}$ stick is at $50\text{ cm}$.
Distance from pivot ($20\text{ cm}$) to center of mass: $\ell = |50 – 20| = 30\text{ cm} = 0.30\text{ m}$.
Radius of gyration about CM for a rod of length $L = 100\text{ cm}$:
$$k_g^2 = \frac{L^2}{12} = \frac{100^2}{12} = \frac{10000}{12} = \frac{2500}{3}\text{ cm}^2 \approx 833.33\text{ cm}^2$$
Equivalent length:
$$L_{\text{eq}} = \ell + \frac{k_g^2}{\ell} = 30 + \frac{2500 / 3}{30} = 30 + \frac{250}{9} = 30 + 27.78 = \mathbf{57.78\text{ cm}}$$
Answer is 57.78 (or 520/9).
Solution 10:
Moment of inertia of solid cylinder: $I = \frac{1}{2} M R^2 = \frac{1}{2} \times 2 \times (0.1)^2 = 0.01\text{ kg}\cdot\text{m}^2$.
Time period of torsional oscillation:
$$T = 2\pi\sqrt{\frac{I}{C}} = 2\pi\sqrt{\frac{0.01}{0.05}} = 2\pi\sqrt{0.2} \approx 2 \times 3.1416 \times 0.4472 = \mathbf{2.81\text{ s}}$$
Answer is 2.81.