Home › Physics › Waves and Oscillations › Compound Pendulum
Waves and Oscillations
Compound Pendulum
A compound pendulum is any rigid body mounted on a horizontal axis of rotation displaced from its center of mass, capable of oscillating freely under gravity. Its time period depends on its moment of inertia about the suspension axis and the distance from the center of mass to that axis.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Basic definition, time period formula, relation with simple pendulum, and center of suspension/oscillation.
PGT extension
Kater's reversible pendulum, condition for maximum/minimum time period, equivalent simple pendulum length, and radius of gyration relations.
Prerequisites
Simple harmonic motion and differential equation of angular oscillations
Moment of inertia and parallel axis theorem
Torque and angular acceleration dynamics for rigid bodies
Definition and Restoring Torque
A rigid body of mass m is suspended about a horizontal frictionless axis O.
The distance from the center of mass G to the axis of suspension O is denoted by l or h.
The restoring torque for a small angular displacement theta is given by ta$u = -mg l \sin(\theta) \approx -mgl \theta$.
Equivalent Length of Simple Pendulum
The time period of a compound pendulum is identical to that of a simple pendulum of length L_eq.
The equivalent length is defined as $L_eq = (I / m h)$, where I is the moment of inertia about the axis of suspension.
Using the parallel axis theorem, $I = I_G + m h^2 = m k_G^2 + m h^2$, where k_G is the radius of gyration about the center of mass.
Interchangeability of Centers of Suspension and Oscillation
The center of oscillation is a point at distance L_eq from the axis of suspension along the line passing through the center of mass.
If the compound pendulum is inverted and suspended from the center of oscillation, its time period remains unchanged.
The distance between the center of suspension and the center of oscillation is always equal to the length of the equivalent simple pendulum.
Conditions for Minimum Time Period
The time period depends on the distance h of the pivot from the center of mass.
Differentiating the time period expression with respect to h reveals that the time period is minimum when h equals the radius of gyration about the center of mass ($h = k_G)$.
The graph of T versus h is symmetrical on both sides of the center of mass.
Derivations
Time Period of a Compound Pendulum
Conditions: Rigid body oscillating with small angular amplitude theta
Write the equation of rotational motion: tau = I alpha, where alph$a = d^2theta / dt^2$.
Equate restoring torque to moment of inertia times angular acceleration: -m g h sin(theta) = I (d^2theta / dt^2).
For small theta, sin(theta) approx theta, yielding (d^2theta / dt^2) + (m g h / I) theta = 0.
Compare with standard SHM equation (d^2theta / dt^2) + omega^2 theta = 0 to get angular frequency omeg$a = \sqrt(m g h / I)$.
Substitute omeg$a = 2 \pi / T$ to obtain the time period $T = 2 \pi \sqrt(I / m g h)$.
Result: T = 2 \pi \sqrt{\frac{I}{m g h}}
Equivalent Simple Pendulum Length and Radius of Gyration
Conditions: Using Parallel Axis Theorem
Express moment of inertia about pivot O using parallel axis theorem: $I = I_G + m h^2 = m k_G^2 + m h^2$, where k_G is radius of gyration about center of mass.
Substitute I into the time period formula: $T = 2 \pi \sqrt((m k_G^2 + m h^2) / m g h)$.
Simplify the expression under the square root: $T = 2 \pi \sqrt((k_G^2 + h^2) / (g h))$.
Compare with simple pendulum formula $T = 2 \pi \sqrt(L_eq / g)$ to define equivalent length.
Result: $L_{eq} = h + \frac{k_G^2}{h}$
Formula sheet
Formula / law
Equation
Variables and units
Conditions
Time Period of Compound Pendulum
$$T = 2 \pi \sqrt{\frac{I}{m g h}}$$
I = Moment of inertia about axis, m = mass, g = acceleration due to gravity, h = distance from pivot to center of mass s
Small amplitude oscillations (theta < 4 degrees)
Equivalent Simple Pendulum Length
$$L_{eq} = \frac{I}{m h} = h + \frac{k_G^2}{h}$$
$k_G = Radius of gyration about center of mass$, h = distance to COM m
Rigid body rotation about a fixed horizontal axis
Moment of Inertia via Parallel Axis Theorem
$$I = I_G + m h^2 = m (k_G^2 + h^2)$$
$I_G = Moment of inertia about center of mass axis$ kg m^2
Axes must be parallel
Condition for Minimum Time Period
$$h = k_G$$
h = distance from pivot, $k_G = radius of gyration$ m
First derivative of T with respect to h set to zero
Minimum Time Period
$$T_{min} = 2 \pi \sqrt{\frac{2 k_G}{g}}$$
$k_G = radius of gyration$, g = acceleration due to gravity s
When $h = k_G$
Frequency of Compound Pendulum
$$f = \frac{1}{2 \pi} \sqrt{\frac{m g h}{I}}$$
f = frequency Hz
Small angle approximation
Relation for Centers of Suspension and Oscillation
$$L_{eq} = l_1 + l_2$$
$l_1 = distance from pivot$ to COM, $l_2 = distance from COM$ to center of oscillation m
In Kater's pendulum configuration
Angular Frequency
$$\omega = \sqrt{\frac{m g h}{I}}$$
omega = angular frequency rad s^{-1}
Small amplitude
Graphs and experiments
Variation of Time Period with Distance from Center of Mass
Draw: A plot of T (y-axis) versus h (x-axis) showing two symmetrical hyperbolic branches with a minimum time period at $h = k_G$.
Interpret: For any given time period other than T_min, there are four points of suspension (two on each side of COM) having the same time period.
Exam use: Frequently tested in PGT to find the radius of gyration or acceleration due to gravity using intercept methods.
Kater's Reversible Pendulum Experiment
Draw: A rigid bar with two fixed knife edges facing each other and two adjustable masses to equalize the time period when suspended from either knife edge.
Interpret: When the time periods about both knife edges are exactly equal, the distance between the knife edges is equal to the length of the equivalent simple pendulum, allowing precise measurement of g.
Exam use: Theoretical questions regarding how Kater's pendulum eliminates errors due to unknown moment of inertia.
Phase Portrait of Compound Pendulum Oscillation
Draw: Angular velocity (omega) versus angular position (theta) showing closed elliptical trajectories for small oscillations.
Interpret: Represents conservative conservation of energy in phase space for the compound pendulum.
Exam use: Conceptual PGT questions on phase space trajectories.
Solved numericals
TGT core
A uniform rod of mass 2 kg and length 3 m is suspended about one of its ends. Find the time period of small oscillations. (Take $g = 9.8 m/s^2)$
Distance of center of mass from pivot $h = L / 2 = 3 / 2 = 1.5 m$.
Moment of inertia of a uniform rod about one end is $I = (1/3) m L^2 = (1/3) * 2 * (3)^2 = 6 kg m^2$.
Use time period formula $T = 2 \pi \sqrt(I / (m g h))$.
A uniform circular disc of mass M and radius R is suspended about a horizontal pin passing through a point on its rim perpendicular to its plane. Calculate its equivalent length of simple pendulum.
Distance from the pivot (edge) to the center of mass is h = R.
Moment of inertia of a disc about its center of mass is $I_G = (1/2) M R^2$.
Using parallel axis theorem, moment of inertia about the edge is $I = I_G + M h^2 = (1/2) M R^2 + M R^2 = (3/2) M R^2$.
Equivalent length is $L_eq = I / (M h) = ((3/2) M R^2) / (M R) = (3/2) R$.
Answer: 1.5 R
PGT extension
Find the minimum possible time period of a uniform thin rod of length 2 m oscillating about a horizontal axis perpendicular to its length. ($g = 9.8 m/s^2)$
Radius of gyration of a rod about its center of mass is given by $k_G^2 = L^2 / 12$.
Given L = 2 m, so $k_G^2 = (2)^2 / 12 = 4 / 12 = 1/3 m^2$. Thus $k_G = 1 / \sqrt(3) m$.
The minimum time period occurs when $h = k_G = 1 / \sqrt(3) m$.
The angular amplitude is assumed to be very small (theta << 1 radian) so that sin(theta) approx theta is valid.
Air resistance and friction at the pivot are neglected for ideal harmonic motion.
The rigid body is assumed not to undergo any elastic deformation during oscillation.
Common misconceptions
Common error: The time period of a compound pendulum depends only on its total mass.
Correct: The time period depends on the ratio of the moment of inertia to the product of mass, gravity, and distance to COM (I / mh), making mass cancel out for uniform bodies.
Mass appears in both the restoring torque and inertial terms, hence cancels out.
Common error: The center of oscillation is always at the geometric lower end of the pendulum.
Correct: The center of oscillation is at a distance L_eq from the pivot, which depends on the moment of inertia and may lie outside the physical body in certain configurations.
L_eq is determined strictly by dynamical parameters I and m h, not physical boundaries.
Common error: Minimum time period occurs when the pivot is at the center of mass.
Correct: When the pivot is at the center of mass, h = 0 and the time period is infinite.
Zero restoring torque results in infinite time period.
Quick revision
Compound pendulum time period formula is $T = 2 \pi \sqrt(I / mgh)$.
Equivalent simple pendulum length is $L_eq = h + k_G^2 / h$.
Minimum time period occurs when $h = k_G$.
Centers of suspension and oscillation are mutually interchangeable.
Must remember
The time period of a compound pendulum is given by $T = 2 \pi \sqrt{\frac{I}{m g h}}$ for small oscillations ($\theta < 4^\circ$).
The equivalent length of a simple pendulum is $L_{eq} = \frac{I}{m h} = h + \frac{k_G^2}{h}$.
The centers of suspension and oscillation are mutually interchangeable without altering the time period.
The time period is minimum when the distance of the pivot from the center of mass equals the radius of gyration ($h = k_G$).
The mass $m$ of the rigid body cancels out in the expression for the time period for uniform bodies.
Glossary
Compound PendulumAny rigid body suspended about a horizontal axis that is free to oscillate in a vertical plane under gravity.
Radius of Gyration ($k_G$)The root-mean-square distance of the mass particles of the body from the axis passing through the center of mass.
Center of OscillationA point situated at a distance $L_{eq}$ from the axis of suspension along the line passing through the center of mass.
Equivalent Length ($L_{eq}$)The length of an equivalent simple pendulum whose time period matches that of the compound pendulum.
Simple Pendulum vs Compound Pendulum
Parameter
Simple Pendulum
Compound Pendulum
Mass Distribution
Concentrated at a single point (idealized bob)
Distributed over a rigid extended body
Moment of Inertia about Pivot
$I = m l^2$
$I = I_G + m h^2$
Time Period Formula
$T = 2 \pi \sqrt{\frac{l}{g}}$
$T = 2 \pi \sqrt{\frac{I}{m g h}}$
Equivalent Length
Equal to length of the string $l$
$L_{eq} = \frac{I}{m h} = h + \frac{k_G^2}{h}$
Practice Quiz
Choose an option to reveal the explanation.
1. What is the formula for the time period $T$ of a compound pendulum?
Explanation: The time period of a compound pendulum is derived from the equation of motion as $T = 2 \pi \sqrt{\frac{I}{m g h}}$.
2. What happens to the time period of a compound pendulum when the pivot is placed exactly at the center of mass ($h = 0$)?
Explanation: When $h = 0$, the restoring torque becomes zero, and the time period tends to infinity ($T \to \infty$).
3. The equivalent length of a simple pendulum for a compound pendulum is given by which expression?
Explanation: The equivalent length is defined as $L_{eq} = \frac{I}{m h}$, where $I$ is the moment of inertia about the pivot and $h$ is the distance to the center of mass.
4. According to the parallel axis theorem, the moment of inertia $I$ about an axis parallel to the axis through the center of mass is:
Explanation: The parallel axis theorem states that $I = I_G + m h^2 = m k_G^2 + m h^2$.
5. Under what condition is the time period of a compound pendulum minimum?
Explanation: Differentiating $T$ with respect to $h$ and setting it to zero reveals that $T$ is minimum when the distance of the pivot from the center of mass equals the radius of gyration ($h = k_G$).
6. What is the minimum time period $T_{min}$ of a compound pendulum in terms of its radius of gyration $k_G$?
Explanation: Substituting $h = k_G$ into the time period formula yields $T_{min} = 2 \pi \sqrt{\frac{m(k_G^2 + k_G^2)}{m g k_G}} = 2 \pi \sqrt{\frac{2 k_G}{g}}$.
7. If a compound pendulum is inverted and suspended from its center of oscillation, its time period will:
Explanation: The centers of suspension and oscillation are interchangeable; suspending the pendulum from the center of oscillation leaves the time period unchanged.
8. The distance between the center of suspension and the center of oscillation is always equal to:
Explanation: By definition, the distance between the center of suspension and the center of oscillation equals the equivalent simple pendulum length $L_{eq}$.
9. What is the angular frequency $\omega$ of a compound pendulum for small oscillations?
Explanation: From the equation of motion $\tau = -mgh\theta = I \ddot{\theta}$, the angular frequency is $\omega = \sqrt{\frac{m g h}{I}}$.
10. Which of the following parameters does NOT affect the time period of a uniform compound pendulum?
Explanation: Mass appears in both the numerator (via moment of inertia) and denominator, causing it to cancel out for uniform bodies.
11. What is the unit of the equivalent length $L_{eq}$ of a compound pendulum?
Explanation: Equivalent length is a length dimension, expressed in meters (m).
12. The graph of time period $T$ versus distance $h$ from the center of mass for a compound pendulum is:
Explanation: The graph of $T$ versus $h$ is symmetrical about the center of mass, showing minimums at $h = \pm k_G$ and tending to infinity at $h = 0$.
13. For a uniform rod of length $l$ suspended about one end, what is the distance $h$ to the center of mass?
Explanation: The center of mass of a uniform rod of length $l$ lies at its geometric center, so the distance from one end is $h = \frac{l}{2}$.
14. What is the small angle approximation condition required for the restoring torque formula $\tau \approx -m g h \theta$?
Explanation: For small amplitudes where $\theta < 4^\circ$, $\sin\theta \approx \theta$, linearizing the restoring torque.
15. Frequency $f$ of a compound pendulum is related to its angular frequency $\omega$ by:
Explanation: Frequency is given by $f = \frac{\omega}{2 \pi} = \frac{1}{2 \pi} \sqrt{\frac{m g h}{I}}$.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. A uniform rod of length $l$ is suspended from one of its ends. What is the time period of small oscillations?
Explanation: For a uniform rod suspended at the end, $I = \frac{1}{3}ml^2$ and $h = \frac{l}{2}$. Thus, $T = 2 \pi \sqrt{\frac{\frac{1}{3}ml^2}{m g (l/2)}} = 2 \pi \sqrt{\frac{2l}{3g}}$ gegen.
2. The radius of gyration of a body about its center of mass is $k_G$. If it is suspended at a distance $h = k_G$ from the center of mass, the minimum time period is:
Explanation: Substituting $h = k_G$ into the time period equation gives $T_{min} = 2 \pi \sqrt{\frac{m(k_G^2 + k_G^2)}{m g k_G}} = 2 \pi \sqrt{\frac{2 k_G}{g}}$.
3. If the length of equivalent simple pendulum for a compound pendulum is $L_{eq}$, what is the relation between the distances of the center of suspension ($l_1$) and center of oscillation ($l_2$) from the center of mass?
Explanation: The equivalent length is the total distance between the center of suspension and center of oscillation, which is $L_{eq} = l_1 + l_2$.
4. A circular disc of mass $m$ and radius $R$ is suspended about a horizontal axis perpendicular to its plane and passing through its rim. What is its time period?
Explanation: For a disc pivoted at the rim, $I_G = \frac{1}{2}mR^2$ and $h = R$. By parallel axis theorem, $I = \frac{1}{2}mR^2 + mR^2 = \frac{3}{2}mR^2$. Thus $T = 2 \pi \sqrt{\frac{3R}{2g}}$.
5. What is the ratio of the minimum time period of a compound pendulum to the time period when the pivot is at distance $h = 2 k_G$?
Explanation: At $h = k_G$, $T_{min} = 2 \pi \sqrt{\frac{2k_G}{g}}$. At $h = 2k_G$, $T = 2 \pi \sqrt{\frac{(k_G^2 + 4k_G^2)}{g(2k_G)}} = 2 \pi \sqrt{\frac{5k_G}{2g}}$. Ratio is $\sqrt{\frac{4}{5}}$... wait, let's recompute: $T_1 \propto \sqrt{2}$, $T_2 \propto \sqrt{\frac{5}{2}}$, ratio is $\sqrt{\frac{4}{5}}$. Let's adjust options or pick correct one matching calculation: $\sqrt{\frac{4}{5}}$. Wait, let's use standard options: $\sqrt{\frac{3}{4}}$ is just a distractor, let's ensure index aligns with correct mathematical ratio.
6. In a physical pendulum, the moment of inertia about the center of mass is $I_G$. If the mass is $m$ and distance to pivot is $h$, the equivalent simple pendulum length is:
Explanation: Using $I = I_G + m h^2$, the equivalent length is $L_{eq} = \frac{I}{m h} = \frac{I_G + m h^2}{m h}$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. Which of the following represents the correct dimension of the radius of gyration $k_G$?
Explanation: Radius of gyration has the dimension of length, i.e., [L].
2. If the distance $h$ of the pivot from the center of mass is increased from $k_G$ to a very large value, the time period $T$ will:
Explanation: As $h \to \infty$, $L_{eq} = h + \frac{k_G^2}{h} \approx h$, and $T = 2 \pi \sqrt{\frac{L_{eq}}{g}} \approx 2 \pi \sqrt{\frac{h}{g}}$, thus $T$ increases as $\sqrt{h}$ for large $h$.
3. A ring of mass $m$ and radius $R$ is suspended from a peg on its circumference. Its time period of oscillation is:
Explanation: For a ring pivoted at its rim, $I_G = mR^2$ and $h = R$. Thus $I = mR^2 + mR^2 = 2mR^2$, and $T = 2 \pi \sqrt{\frac{2mR^2}{mgR}} = 2 \pi \sqrt{\frac{2R}{g}}$.
4. What is the physical significance of the center of oscillation in a compound pendulum?
Explanation: The center of oscillation acts as the point where the entire mass can be considered concentrated to form an equivalent simple pendulum of length $L_{eq}$.
5. If the time period of a compound pendulum is $T$, what is the formula for frequency $f$?
Explanation: Frequency is the reciprocal of the time period: $f = \frac{1}{T} = \frac{1}{2 \pi} \sqrt{\frac{m g h}{I}}$.
6. In Kater's pendulum, two knife edges are used to:
Explanation: Kater's pendulum utilizes two reversible knife edges to adjust positions so that the time period is identical, allowing precise measurement of acceleration due to gravity $g$.
7. If a rigid body has two points of suspension about which the time period is the same, then the distance between these two points is equal to:
Explanation: The distance between the center of suspension and the center of oscillation (which acts as the second suspension point with the same period) is equal to $L_{eq}$.
8. How does the time period of a compound pendulum change if the acceleration due to gravity $g$ increases (e.g., moving from equator to poles)?
Explanation: Since $T$ is inversely proportional to $\sqrt{g}$, an increase in $g$ results in a decrease in the time period $T$.
9. For a uniform square lamina of side $a$ suspended about a horizontal axis through one of its corners parallel to its plane, what is $h$?
Explanation: The distance from a corner of a square lamina of side $a$ to its center of mass is half the diagonal, which is $h = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.
10. Which theorem is indispensable in finding the moment of inertia of a rigid body about an arbitrary parallel axis passing through the pivot?
Explanation: The parallel axis theorem ($I = I_G + mh^2$) is used to shift the axis of rotation from the center of mass to the pivot point.