Experiment: simple pendulum and loss of energy

What is this experiment?

In practice, oscillating bodies eventually come to rest at their equilibrium positions because of the damping due to friction and other dissipative causes.

NCERT Physics Part II (Class 11), Section 13.2, page 260

Both the JEE Main 2027 and the NEET (UG) 2026 syllabuses list this under Unit 20, Experimental Skills: “Simple pendulum-dissipation of energy by plotting a graph between the square of amplitude and time”. The NCERT textbook derives the period (Section 13.8); NCERT’s Class XI Physics Laboratory Manual gives the method in Experiment 6 (pages 60–67) and Activity 7 (pages 148–151). This page adapts those methods in two parts: measuring the time period, and following the square of the amplitude with time.

Apparatus

• A heavy metal bob with a hook, on a light, inextensible thread about 1 m long.
• A rigid clamp stand with a split cork, so that the thread swings from a fixed point.
• A stopwatch reading to 0.1 s (or better); a balance if absolute energy, rather than only relative energy loss, is required.
• A metre scale, to measure the length, and a half-metre scale fixed horizontally just below the bob, centred on its rest position, to read the amplitude.
• Vernier callipers, to measure the diameter of the bob.

Theory

Period. For small swings the motion is simple harmonic with T = 2π√(L/g) (NCERT Eq. 13.26, Section 13.8; see the derivation for the simple pendulum). Here L is the effective length, from the point of suspension to the centre of the bob: include the hook and bob radius according to the points used when measuring the thread. Rearranging, g = 4π^²L/T^².

Energy. A simple harmonic oscillator has total energy E = ½mω^²A^², where A is its amplitude (NCERT Eq. 13.18 with k = mω^²). For a pendulum ω^² = g/L, so for small swings, with A the sideways displacement of the bob,

For a given pendulum, m, g and L are fixed, so the energy is proportional to A^². A graph of A^² against time is therefore a graph of the energy against time, up to a constant factor. (The same result follows from E = mgh, since the bob rises by h ≈ A^²/(2L) at the end of each swing.)

Why the energy falls. Air resistance on the bob and thread, and friction at the point of suspension, do negative work on every swing, so the mechanical energy is not conserved: it is turned into heat. The simplest model (beyond the NCERT text) takes the drag as proportional to the speed, like viscous drag. For small underdamped oscillations with linear drag, successive turning-point amplitudes follow an exponential envelope. Their squared values, and hence turning-point energies, fall by the same fraction over equal numbers of cycles. Plot ln(A^²/A_₀^²) against time to test this; instantaneous energy need not be an exact pure exponential within a cycle. For a bob a few centimetres across, moving at tens of centimetres per second, real air drag grows faster than the speed (closer to v^²), so a real A^²–t curve falls steeply at first and more gently later but need not be exactly exponential.

Part 1: measuring the time period

1. Measure the length of the thread from the point of suspension to the hook, and the diameter of the bob with vernier callipers. L = thread length + hook length + radius of the bob.
2. Fix a mark (a vertical line or the rest position on the scale) behind the bob when it hangs at rest.
3. Pull the bob aside by a small amount, about 5 cm for a 1 m pendulum (a few degrees), and release it without pushing, so it swings in one plane.
4. Start the stopwatch as the bob passes the mean position, and count the oscillations, starting the count at zero. Count each return through the mark in the same direction as one complete oscillation, and stop after 20 (or more).
5. Repeat and take the mean time t. The period is T = t/20.
6. Repeat for several lengths. A graph of L against T^² is a straight line through the origin with slope g/4π^². If you plot the thread length alone, forgetting the radius r of the bob, a constant omitted length δ (bob radius plus any unmeasured hook) shifts the length-axis intercept to −δ without changing the ideal slope. Fit an intercept rather than forcing the line through zero; length-dependent errors can still bias the slope. This is one reason to find g from the slope.

Why time many oscillations? If the absolute start/stop timing uncertainty is similar for both runs, dividing a 20-cycle time by 20 reduces its contribution to the uncertainty of T by a factor of 20 (Example 1).

Part 2: square of the amplitude against time

1. Fix the half-metre scale horizontally just below the bob, with its middle under the rest position.
2. Release the bob gently and let a few swings settle into one plane. At a chosen turning point, record the initial amplitude (for example 10 cm) and set that time as zero.
3. Every chosen number of complete oscillations, read the turning-point displacement on the same side and record its time. NCERT suggests every 5 or 10 cycles; the illustrative table below uses 20.
4. Continue until the amplitude is too small to read well.
5. Tabulate t, A and A^², and plot A^² against t.

A sample observation table (illustrative values, used in Example 2; L = 1.0 m, T ≈ 2.0 s):

Result. The graph of A^² against t is a curve that falls steadily: the energy of the pendulum decreases with time. In these illustrative values it falls by about the same fraction (to about 0.79 of its value) in every 40 s; the amplitude data alone cannot identify whether air drag, support friction or other losses dominate.

Precautions. Use a heavy, compact bob, so that air resistance has little effect, and a light, inextensible thread, so that its mass can be ignored and the length stays fixed. Clamp the thread firmly between the halves of a split cork, so the point of suspension does not move. Keep the amplitude small, so the motion is simple harmonic. Release the bob without a push, and make sure it swings in a plane and does not spin or move in an ellipse. Read the amplitude with your eye directly above the bob to avoid parallax, and keep draughts away. Count oscillations from the mean position, where the bob moves fastest and its passage is easiest to time.

Sources of error. Reaction time in starting and stopping the stopwatch; parallax and the moving bob when reading the amplitude; uncertainty in the position of the centre of the bob when measuring L. For g = 4π^²L/T^², a first-order worst-case uncertainty estimate is Δg/g ≈ ΔL/L + 2ΔT/T, using positive uncertainty magnitudes, as in the Young’s modulus experiment.

Solved examples

Common mistakes to avoid

Quick revision summary

• Period: time n oscillations, T = t/n; then g = 4π^²L/T^² with L to the centre of the bob.
• L against T^² is a straight line through the origin with slope g/4π^².
• Energy of a pendulum: E = mgA^²/(2L), proportional to A^².
• Reading A every few oscillations and plotting A^² against t shows the energy falling, because of air resistance and friction at the support.
• The illustrative turning-point data are approximately exponential. Linear underdamped motion predicts an exponential amplitude envelope; other damping laws need not.
• First-order worst-case estimate: Δg/g ≈ ΔL/L + 2ΔT/T. Independent standard uncertainties instead combine in quadrature.

Questions students ask

What does dissipation of energy mean in the simple pendulum experiment?

In the simple pendulum experiment, dissipation of energy means that the pendulum’s mechanical energy is gradually turned into heat, the random internal energy of the air and of the support, which does not spontaneously reassemble into the original coherent swing. The energy is not destroyed, so energy as a whole is conserved, but mechanical energy is not: each swing hands a little more of it to the surroundings. That is why a free pendulum eventually stops, and why the square of the amplitude, which is proportional to the pendulum’s energy, falls with time.

Does the time period of a pendulum change as its amplitude dies away?

For small swings and weak damping, the period is approximately constant as the amplitude decays: T = 2π√(L/g) does not depend on the amplitude, The linear damped frequency is slightly below the undamped value; whether the difference is detectable depends on the damping strength and timing precision. Early and late timing runs should agree within their uncertainties when this model applies. Larger swings take slightly longer, about 0.8% longer at 20° and 4% at 45°, so the period of a pendulum started at a large angle shortens slightly as its swings shrink.

Why does a heavy bob keep swinging longer than a light bob of the same size?

If air drag dominates and shape, size, initial speed and length are the same, a heavier bob generally loses a smaller fraction of its energy per swing because the air resistance on a bob depends on its size, shape and speed, not on its mass, while its energy, E = mgA^²/(2L), is proportional to its mass. So with the same drag a heavy bob loses a smaller fraction of its energy in each swing, and its amplitude falls more slowly: a denser ball can therefore keep swinging longer. Support friction may also depend on load, so this comparison is not universal. That is why the precautions ask for a heavy, compact bob: air resistance then has little effect on each swing, and the swings last long enough to follow.

What is the shape of the graph between length and time period of a simple pendulum?

The graph of length L against time period T of a simple pendulum is a parabola, because L = (g/4π^²)T^²: doubling the period needs four times the length. Drawn the other way, T against L, it is the same parabola on its side, rising steeply at first and then more slowly. Plotting L against T^² instead gives a straight line through the origin, which is why that graph is the one used to find g.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Why is the stopwatch started and stopped as the bob passes the mean position, not at an extreme?

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   At the mean position the bob moves fastest, so the moment it passes a mark is sharp and easy to judge. At an extreme it moves slowly and seems to hang there, so the moment of turning is hard to pick.

2. Q2. A pendulum has a thread 0.62 m long and a bob of diameter 4 cm. Find its effective length and expected period (g = 9.8 m s^⁻²).

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   L = 0.62 + 0.02 = 0.64 m. T = 2π√(0.64/9.8), about 1.61 s.

3. Q3. What two causes make the amplitude of a real pendulum decrease?

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   Air resistance on the bob and thread, and friction at the point of suspension. Both turn some of the pendulum’s mechanical energy into heat on every swing.