Simple Pendulum

What a simple pendulum is

A simple pendulum is an idealisation: a small bob of mass $m$, treated as a point mass, tied to an inextensible, massless string of length $L$, the other end of which is fixed to a rigid support. The length $L$ is measured from the point of suspension to the centre of gravity of the bob. Galileo is said to have first noticed the regularity of such motion by timing a swinging chandelier in a church against his own pulse.

When the bob is pulled aside by a small angle and released, it executes angular oscillations in a vertical plane about the lowest point, which is its mean (equilibrium) position. At the mean position the angle $\theta$ that the string makes with the vertical is zero. The pendulum is the gravitational cousin of the spring–mass system — but here the restoring agency is not a spring constant, it is gravity itself acting through the geometry of the swing.

The restoring torque on the bob

Only two forces act on the bob: the tension $T$ along the string and the weight $mg$ acting vertically downward. NCERT resolves the weight into a component $mg\cos\theta$ along the string and a component $mg\sin\theta$ perpendicular to it. Because the bob moves along a circular arc of radius $L$ centred on the support, it has a radial acceleration $\omega^2 L$ and a tangential acceleration. The net radial force $T - mg\cos\theta$ supplies the centripetal requirement, while the tangential acceleration is supplied entirely by $mg\sin\theta$.

NCERT chooses to work with torque about the support, because the radial force passes through the support and therefore exerts zero torque there. The torque is provided only by the tangential component:

$$\tau = -L\,(mg\sin\theta) \tag{NCERT 13.19}$$

The negative sign marks this as a restoring torque — it always acts to reduce the angular displacement and drag the bob back toward the mean position. Applying Newton's law of rotational motion $\tau = I\alpha$, where $I$ is the moment of inertia of the bob about the support and $\alpha$ is the angular acceleration:

$$I\alpha = -mg\sin\theta\,L \quad\Rightarrow\quad \alpha = -\frac{mgL}{I}\sin\theta \tag{NCERT 13.22}$$

The small-angle approximation

Equation 13.22 is exact, but it is not the SHM equation, because $\alpha$ is proportional to $\sin\theta$ rather than to $\theta$. To recover simple harmonic motion we use the series expansion of the sine, with $\theta$ in radians:

$$\sin\theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots \tag{NCERT 13.23}$$

If $\theta$ is small, the cubic and higher terms are negligible and $\sin\theta \approx \theta$. NCERT's Table 13.1 tabulates this: even for an angle as large as $20^\circ$, the value of $\sin\theta$ is nearly identical to $\theta$ expressed in radians. In practice "small" means roughly $\theta \lesssim 10^\circ$ for high accuracy, stretching to about $20^\circ$ before the error becomes noticeable.

With the approximation in place, Equation 13.22 becomes

$$\alpha = -\frac{mgL}{I}\,\theta \tag{NCERT 13.24}$$

This has exactly the structure of the SHM equation $a = -\omega^2 x$, with angular displacement $\theta$ playing the role of $x$. Compare it with the force law for SHM: the restoring effect is proportional to the displacement and directed toward equilibrium. The pendulum is therefore simple harmonic for small angles, and not otherwise.

NCERT derivation of T = 2π√(L/g)

Comparing Equation 13.24 with the standard SHM relation $\alpha = -\omega^2\theta$ identifies the angular frequency, and hence the period, in terms of the moment of inertia:

$$\omega = \sqrt{\frac{mgL}{I}} \qquad\text{and}\qquad T = 2\pi\sqrt{\frac{I}{mgL}} \tag{NCERT 13.25}$$

The final step uses the fact that the string is massless, so the only contribution to the moment of inertia about the support is the point bob at distance $L$: $I = mL^2$. Substituting:

$$T = 2\pi\sqrt{\frac{mL^2}{mgL}} = 2\pi\sqrt{\frac{L}{g}} \tag{NCERT 13.26}$$

The mass cancels in this last line, and the result is the well-known formula for the period of a simple pendulum. It is worth keeping the NIOS route in mind as a cross-check: NIOS treats $mg\sin\theta$ as the restoring force directly, writes $F = mg\theta = mg\,x/L$ for small displacement $x$ along the arc, identifies a force-per-displacement $k = mg/L$, and obtains $\omega = \sqrt{g/L}$ — the same answer by a force argument rather than a torque argument.

Independence of mass and amplitude

Two properties of $T = 2\pi\sqrt{L/g}$ are tested far more often than the derivation itself.

Independence of mass. The formula contains no $m$. Physically, the restoring torque $-L\,mg\sin\theta$ and the inertia $I = mL^2$ both scale with $m$, so the mass cancels exactly. A heavy iron bob and a light cork bob on equal-length strings keep identical time. This is the deepest reason a pendulum clock does not need recalibration if you swap the bob for a heavier one of the same length.

Independence of amplitude — isochronism. Within the small-angle regime the period depends only on $L$ and $g$, not on how far the bob is pulled aside. Doubling a small amplitude leaves $T$ unchanged. This is the isochronism Galileo observed, and it is what makes the pendulum a viable timekeeper. The independence is approximate: at genuinely large angles the period creeps upward with amplitude.

What the period actually depends on

Because $T = 2\pi\sqrt{L/g}$, the period scales as the square root of length and inversely as the square root of gravity. The four factors a NEET question can manipulate are summarised below.

Effective gravity — lifts, depth, planets

The single most productive idea for NEET pendulum problems is that $g$ in the formula is whatever effective gravitational acceleration the bob actually feels in its frame. Replace $g$ by $g_{\text{eff}}$ and the formula does the rest:

$$T = 2\pi\sqrt{\frac{L}{g_{\text{eff}}}}$$

In a lift accelerating upward with acceleration $a$, the bob feels a stronger pull, $g_{\text{eff}} = g + a$, so the pendulum runs faster (shorter period). In a lift accelerating downward with acceleration $a$, $g_{\text{eff}} = g - a$ and the period lengthens. The extreme case is free fall, where $a = g$, giving $g_{\text{eff}} = 0$: the string goes slack, there is no restoring agency, and the pendulum simply does not oscillate — its period is effectively infinite. The same logic covers a different planet, where $g$ is set by that planet's surface gravity.

The seconds pendulum

A seconds pendulum is one whose time period is exactly $T = 2$ s, so each one-way swing from one extreme to the mean takes one second. Inverting the formula gives its length:

$$L = \frac{gT^2}{4\pi^2}$$