Chapter 13 · Oscillations

13.3 The Spring–Block Oscillator and the Simple Pendulum

Apply the force law to the two classic oscillators: a block on a spring, $T=2\pi\sqrt{m/k}$ (horizontal or hanging vertically), and a simple pendulum swinging through a small angle, $T=2\pi\sqrt{l/g}$ — the basis of pendulum clocks and of measuring $g$.

How do you weigh an astronaut on the International Space Station? A bathroom scale is useless — everything on board is in free fall, and the astronaut presses on the scale with no force at all. Mass, however, has not gone anywhere: a heavier astronaut is harder to accelerate. So the crew sit on a chair mounted on springs and set it oscillating. The period tells them the mass, through exactly the formula of this section. On the ground, the same physics times the bounce of a car on its suspension, and its partner — the pendulum — kept the world's time for nearly three hundred years and still measures the strength of gravity in geophysical surveys.

This section applies the force law of Section 13.2 to the two systems every laboratory owns. For a block on a spring the restoring force is exactly linear, and the period is $T=2\pi\sqrt{m/k}$, whether the spring lies horizontally or hangs vertically. For a simple pendulum the restoring force is linear only for small swings; then $T=2\pi\sqrt{l/g}$, which depends on neither the mass of the bob nor the size of the swing, and turns a piece of string and a stopwatch into an instrument for measuring $g$.

The spring–block oscillator

A block of mass $m$ rests on a smooth horizontal surface and is attached to an ideal, light spring of spring constant $k$ whose other end is fixed to a wall. With the spring at its natural length the block is in equilibrium; take this position as $x=0$. Pull the block out to $x$ and the spring pulls back with $F=-kx$ (Hooke's law, Definition 5.6 and Section 8.1). The weight and the normal force cancel, and there is no friction, so this is the net force: a linear restoring force. By Result 13.5 the block moves in SHM.

📐 Result 13.7 — Period of a spring–block oscillator

A block of mass $m$ attached to a light spring of spring constant $k$, on a smooth surface, oscillates in SHM with

$$\omega=\sqrt{\frac km},\qquad T=2\pi\sqrt{\frac mk},\qquad \nu=\frac1{2\pi}\sqrt{\frac km},$$

for any amplitude within the spring's elastic (Hooke's-law) range. The period does not depend on the amplitude or on $g$; it is longer for a heavier block and shorter for a stiffer spring.

The formula assumes a massless spring and no friction. A real spring's own mass also moves, which lengthens the period slightly; friction makes the oscillation die away (see the extension box). Because $T$ depends on $m$ but not on $g$, a spring oscillator measures mass — the inertia — even where there is no weight.

✏️ Example 1 — Weighing an astronaut
A body-mass measuring chair on a space station is attached to a spring system of effective spring constant $600\ \text{N/m}$. The empty chair has a mass of $12.0\ \text{kg}$. With an astronaut strapped in, the chair oscillates with a period of $2.20\ \text{s}$. (a) Find the astronaut's mass. (b) Find the period of the empty chair. (c) Why would a pendulum be useless here?
(a) $T=2\pi\sqrt{\dfrac Mk}$ gives $M=\dfrac{kT^2}{4\pi^2}=\dfrac{600\times2.20^2}{4\pi^2}=73.6\ \text{kg}$ for chair plus astronaut, so the astronaut's mass is $73.6-12.0=61.6\ \text{kg}$.
(b) $T_0=2\pi\sqrt{\dfrac{12.0}{600}}=0.889\ \text{s}$.
(c) A pendulum's period depends on $g$, and in orbiting free fall the effective $g$ is zero: a pendulum would not swing back at all. The spring's force does not depend on gravity.
Check $\dfrac{T}{T_0}=\sqrt{\dfrac{73.6}{12.0}}=2.48=\dfrac{2.20}{0.889}$ ✓.

A vertical spring. Hang the block from the spring instead. It now sits in equilibrium with the spring stretched by $l_0$, where the spring force balances the weight: $kl_0=mg$. Measure the displacement $y$ downward from this new equilibrium position. The spring is then stretched by $l_0+y$, and the net downward force is $mg-k(l_0+y)=-ky$. The weight has been cancelled exactly by part of the spring force: what remains is again a linear restoring force with the same $k$.

📐 Result 13.8 — Vertical spring–block oscillator

A block hanging from a light spring oscillates in SHM about its equilibrium position (where the spring is stretched by $l_0=\dfrac{mg}{k}$) with the same period as on a horizontal surface:

$$T=2\pi\sqrt{\frac mk}=2\pi\sqrt{\frac{l_0}{g}} .$$

Gravity shifts the mean position but does not change the period.

✏️ Example 2 — A block hanging from a spring
A spring stretches by $9.8\ \text{cm}$ when a $0.50\ \text{kg}$ block is hung from it at rest. The block is pulled down a further $3.0\ \text{cm}$ and released. Find (a) the spring constant; (b) the period; (c) the greatest and least tensions in the spring during the motion; (d) the block's maximum speed.
(a) $k=\dfrac{mg}{l_0}=\dfrac{0.50\times9.8}{0.098}=50\ \text{N/m}$.
(b) $T=2\pi\sqrt{\dfrac{0.098}{9.8}}=2\pi\times0.10=0.63\ \text{s}$ (or $2\pi\sqrt{0.50/50}$, the same).
(c) The amplitude is $A=3.0\ \text{cm}$ about the equilibrium position, so the stretch varies from $9.8-3.0=6.8\ \text{cm}$ to $12.8\ \text{cm}$. Tensions: $50\times0.128=6.4\ \text{N}$ (lowest point) and $50\times0.068=3.4\ \text{N}$ (highest point).
(d) $\omega=10\ \text{rad/s}$: $v_\text{max}=\omega A=0.30\ \text{m/s}$, at the equilibrium position.
Check The tensions are $mg\pm kA=4.9\pm1.5\ \text{N}$, averaging the weight ✓. If the block were pulled down more than $9.8\ \text{cm}$, the spring would go slack at the top and the motion would no longer be SHM.
Interactive 1 — Spring–Block Lab

A block on a smooth surface is pulled $5.0\ \text{cm}$ from equilibrium and released. Its motion is found step by step from Newton's second law with the net spring force, and the period is timed from the simulation. The bars show the kinetic, elastic potential and total energy. Change $m$ and $k_1$ and compare the timed period with $2\pi\sqrt{m/k}$. (The two-spring settings belong to the 🚀 box below.)

2.0 kg100 N/m300 N/m
$k_\text{eff}$—
$2\pi\sqrt{m/k_\text{eff}}$—
timed period—
total energy $E$—

The simple pendulum

📘 Definition 13.5 — Simple pendulum

A simple pendulum is an idealised pendulum: a point mass $m$ (the bob) hanging from a fixed support by a light, inextensible string of length $l$, free to swing in a vertical plane without friction or air resistance. Its displacement is the angle $\theta$ between the string and the vertical.

A small heavy ball on a long thin thread is a good approximation; $l$ is then measured from the support to the centre of the ball. Displace the bob through an angle $\theta$ and release it. Two forces act: the tension $T_\text{s}$ along the string, which passes through the support, and the weight $mg$. Take torques about the support (Section 6.2): the tension has no torque, and the weight has lever arm $l\sin\theta$, so

$$\tau=-mgl\sin\theta,$$

with the minus sign because the torque always turns the bob back towards $\theta=0$. The moment of inertia of a point mass at distance $l$ is $I=ml^2$, so $\tau=I\alpha$ (Section 6.3) gives the angular acceleration

$$\alpha=-\frac gl\sin\theta .$$

This restoring torque is proportional to $\sin\theta$, not to $\theta$, so the motion is not exactly simple harmonic. But for small angles in radians, $\sin\theta\approx\theta$:

$\theta$$5^\circ$$10^\circ$$15^\circ$$20^\circ$
$\theta$ (rad)$0.0873$$0.1745$$0.2618$$0.3491$
$\sin\theta$$0.0872$$0.1736$$0.2588$$0.3420$
error of $\sin\theta\approx\theta$$0.13\%$$0.51\%$$1.2\%$$2.1\%$

Within about $10^\circ$ the approximation is better than about half a per cent, and $\alpha=-\dfrac gl\theta$ — angular acceleration proportional to angular displacement, and opposite to it. This is the angular form of the SHM condition $a=-\omega^2x$, with $\omega^2=\dfrac gl$. (The same follows from forces: the component of the weight along the arc, $-mg\sin\theta\approx-mg\theta=-\dfrac{mg}{l}s$, is a linear restoring force on the arc length $s=l\theta$, with $k=\dfrac{mg}{l}$.)

📐 Result 13.9 — Period of a simple pendulum

For small amplitudes ($\theta_0\lesssim10^\circ$) a simple pendulum of length $l$ executes angular SHM, $\theta=\theta_0\cos(\omega t+\phi)$, with

$$\omega=\sqrt{\frac gl},\qquad T=2\pi\sqrt{\frac lg}.$$

The period is independent of the mass of the bob and, for small swings, of the amplitude; it depends only on the length and on $g$.

The mass cancels because the weight that pulls the bob back and the inertia that resists the motion are both proportional to $m$ — the same reason all bodies fall with the same acceleration. Independence of amplitude (Galileo's observation, called isochronism) is what makes a pendulum a clock: as friction slowly reduces the swing, the period stays the same. For larger amplitudes the period grows: a numerical solution of $\alpha=-\dfrac gl\sin\theta$ gives about $0.8\%$ longer at $20^\circ$, $1.7\%$ at $30^\circ$ and $7.3\%$ at $60^\circ$ (Interactive 2 shows this). A pendulum whose period is exactly $2\ \text{s}$ — one second for each swing — is called a seconds pendulum; its length is $l=\dfrac{gT^2}{4\pi^2}=\dfrac{9.8}{\pi^2}=0.993\ \text{m}$.

✏️ Example 3 — Periods of pendulums
(a) Find the period of a simple pendulum of length $1.20\ \text{m}$. (b) Find the length of a seconds pendulum where $g=9.80\ \text{m/s}^2$. (c) The seconds pendulum is taken to the Moon, where $g=1.62\ \text{m/s}^2$. Find its period there. (d) How would the period in (a) change if the bob's mass were doubled? If the length were doubled?
(a) $T=2\pi\sqrt{\dfrac{1.20}{9.8}}=2.20\ \text{s}$.
(b) $l=\dfrac{gT^2}{4\pi^2}=\dfrac{9.80\times4}{4\pi^2}=0.993\ \text{m}$.
(c) $T\propto\dfrac1{\sqrt g}$: $T=2.00\sqrt{\dfrac{9.80}{1.62}}=4.92\ \text{s}$.
(d) Mass doubled: no change. Length doubled: $T$ increases by $\sqrt2$, to $3.11\ \text{s}$.
Check Dimensions: $\sqrt{l/g}$ has $\sqrt{\text{m}/(\text{m s}^{-2})}=\text{s}$ ✓ — the same combination that dimensional analysis predicts (Section 1.3).

Measuring $g$. Squaring the period formula gives $T^2=\dfrac{4\pi^2}{g}\,l$: a graph of $T^2$ against $l$ is a straight line through the origin with slope $\dfrac{4\pi^2}{g}$. Measuring $T$ for several lengths and drawing the best straight line averages out the random errors in individual timings, and an error in where the length is measured from shows up as a line that misses the origin instead of spoiling the slope.

✏️ Example 4 — $g$ from a $T^2$–$l$ graph
A student times $20$ oscillations of a pendulum (amplitude about $5^\circ$) for four lengths: $l=0.400,\ 0.600,\ 0.800,\ 1.000\ \text{m}$ give times of $25.4,\ 31.1,\ 35.9,\ 40.1\ \text{s}$. Find $T^2$ for each length, the slope of the $T^2$–$l$ graph, and $g$.
Periods $T=\dfrac{t}{20}$: $1.270,\ 1.555,\ 1.795,\ 2.005\ \text{s}$; $T^2=1.613,\ 2.418,\ 3.222,\ 4.020\ \text{s}^2$.
Slope The ratios $\dfrac{T^2}{l}$ are $4.03,\ 4.03,\ 4.03,\ 4.02\ \text{s}^2/\text{m}$: the points lie close to a line through the origin, whose slope is their mean, $4.03\ \text{s}^2/\text{m}$ (from the end points, $\dfrac{4.020-1.613}{1.000-0.400}=4.01$).
g $g=\dfrac{4\pi^2}{\text{slope}}=\dfrac{39.48}{4.03}=9.80\ \text{m/s}^2$.
Check The timings are to $0.1\ \text{s}$ in about $30\ \text{s}$, roughly $0.3\%$, so $T^2$ is good to about $0.6\%$ and $g$ to about $\pm0.06\ \text{m/s}^2$. Interactive 2 lets you build such a graph and see what a large amplitude does to it.
✏️ Example 5 — A pendulum clock on a hill station
A pendulum clock keeps correct time at sea level. It is taken to a hill station $2.0\ \text{km}$ higher. (a) Does it gain or lose time? (b) By how many seconds per day? (c) By how much must its seconds pendulum be shortened to correct it? Take $R_E=6.37\times10^6\ \text{m}$.
(a) At height $h$, $g_h=\dfrac{g}{(1+h/R_E)^2}$ (Section 7.2) is smaller, so $T=2\pi\sqrt{l/g_h}$ is longer: the pendulum swings more slowly and the clock loses time.
(b) $T\propto\dfrac1{\sqrt g}$, so $\dfrac{T_h}{T}=1+\dfrac h{R_E}$ exactly. Fractional loss $\dfrac{2000}{6.37\times10^6}=3.14\times10^{-4}$; in a day, $86\,400\times3.14\times10^{-4}=27\ \text{s}$.
(c) Keep $\dfrac lg$ fixed: $l$ must fall in proportion to $g$, by the fraction $1-\dfrac1{(1+h/R_E)^2}\approx\dfrac{2h}{R_E}=6.28\times10^{-4}$. For $l=0.993\ \text{m}$: shorten by $0.62\ \text{mm}$.
Check A tiny change of $g$ gives a large error in a clock because the error adds up over $86\,400$ seconds; this is why pendulum clocks have a fine adjusting nut under the bob.
Interactive 2 — Pendulum Lab

The solid bob is a true pendulum, found step by step from $\alpha=-\dfrac gl\sin\theta$; the faint bob follows the small-angle SHM $\theta_0\cos\omega t$. At $5^\circ$ they stay together; at $60^\circ$ the true pendulum falls behind. Press Record to add the timed period for the current length to the $T^2$–$l$ graph, then change the length and record again: the slope gives $g$. Try recording with a large amplitude and see the measured $g$ come out too small.

1.00 m5°
small-angle $2\pi\sqrt{l/g}$—
true period (numerical)—
difference—
$g$ from graph slope—
⚠️ Common mistakes

1. $T=2\pi\sqrt{k/m}$. It is $\sqrt{m/k}$: a heavier block oscillates more slowly. Check the units — $\sqrt{\text{kg}/(\text{N/m})}=\text{s}$.

2. Frequency or angular frequency? $\omega=\sqrt{k/m}$ is in rad/s; the frequency is $\nu=\dfrac{\omega}{2\pi}$ in Hz. A $1.0\ \text{kg}$ block on a $100\ \text{N/m}$ spring has $\omega=10\ \text{rad/s}$ but $\nu=1.6\ \text{Hz}$, not $10\ \text{Hz}$.

3. Gravity and the vertical spring. Hanging the spring changes the equilibrium position, not the period. Measure the displacement from the new equilibrium.

4. The pendulum's mass. A heavier bob does not swing faster or slower (for a simple pendulum, neglecting air resistance).

5. Degrees in $\sin\theta\approx\theta$. The approximation holds with $\theta$ in radians, and only for small angles; $T=2\pi\sqrt{l/g}$ is not valid for large swings.

6. Length of the pendulum. $l$ is measured to the centre of the bob, not to its top or bottom.

🚀 Beyond the board syllabus — springs in combination (JEE Main · NEET · JEE Advanced); damped and forced oscillations (JEE Advanced)

Springs in series and parallel are not in the CBSE syllabus or the NCERT chapter, and no 2026 entrance syllabus names them; they are routine problems under "oscillations of a spring" (JEE Main, NEET) and linear SHM (JEE Advanced). Damped and forced oscillations and resonance are listed in JEE Advanced 2026 only — not in CBSE, JEE Main or NEET.

Springs in combination. A car rests on four springs, a mattress on hundreds. Replace the combination by one spring with the effective spring constant $k_\text{eff}$ — the spring that gives the same restoring force for the same displacement, $F=-k_\text{eff}\,x$ — and the block oscillates with $T=2\pi\sqrt{m/k_\text{eff}}$. In parallel (side by side) the displacement $x$ stretches both springs by $x$ and their forces add. In series (end to end) the massless junction must have zero net force, so both springs carry the same tension $F$ and their extensions $\dfrac F{k_1}$, $\dfrac F{k_2}$ add up to $x$. A block between two springs fixed to opposite walls behaves like the parallel case.

$$\text{parallel:}\quad k_\text{p}=k_1+k_2$$ $$\text{series:}\quad \frac1{k_\text{s}}=\frac1{k_1}+\frac1{k_2}$$

Parallel is stiffer than either spring, series softer than either. Cutting a spring: a uniform spring is its pieces in series, so a piece $\dfrac1n$ as long has constant $nk$ ($kL$ is constant — like a wire's stiffness $\dfrac{YA}{L}$, Section 8.2). Interactive 1 has these settings.

Example. Springs of $100\ \text{N/m}$ and $300\ \text{N/m}$ hold a $2.0\ \text{kg}$ block. In parallel, $k_\text{p}=400\ \text{N/m}$ and $T=2\pi\sqrt{2.0/400}=0.44\ \text{s}$; in series, $k_\text{s}=\dfrac{100\times300}{400}=75\ \text{N/m}$ and $T=2\pi\sqrt{2.0/75}=1.03\ \text{s}$ — $\sqrt{400/75}=2.3$ times longer. One spring on each side gives $400\ \text{N/m}$ again: $0.44\ \text{s}$.

Damped oscillations. A real oscillator loses energy to air resistance and internal friction, so its amplitude dies away: the motion is damped. If the damping force is proportional to the velocity, $F_\text{d}=-bv$, the amplitude decays exponentially, $A(t)=A_0e^{-bt/2m}$, while the period stays almost the same as without damping (for light damping). The energy, proportional to $A^2$, decays as $e^{-bt/m}$. Car shock absorbers are deliberately heavy dampers, so that a car body returns to rest after a bump instead of bouncing for a long time.

Forced oscillations and resonance. If an oscillator of natural angular frequency $\omega_0=\sqrt{k/m}$ is pushed by a periodic force of angular frequency $\omega_\text{d}$, after a while it oscillates at the driving frequency $\omega_\text{d}$, with an amplitude that depends on how close $\omega_\text{d}$ is to $\omega_0$. When they match, even a small driving force builds up a large amplitude — resonance. The lighter the damping, the taller and narrower the resonance peak. You push a swing at its natural frequency to make it go high; a radio is tuned by adjusting a circuit's natural frequency to that of the wanted station; soldiers break step when crossing a bridge; and in 2000 London's Millennium Bridge swayed alarmingly when the footsteps of the crowd fell into step with its sideways sway.

Example. A $0.50\ \text{kg}$ block on a spring is damped by a force $-bv$ with $b=0.10\ \text{kg/s}$. Its amplitude halves when $e^{-bt/2m}=\tfrac12$: $t=\dfrac{2m\ln2}{b}=6.9\ \text{s}$.

Exercises

Take $g=9.8\ \text{m/s}^2$ unless stated otherwise. Springs are light and obey Hooke's law; surfaces are smooth; pendulums are simple and swing through small angles.

I
The spring–block oscillator.
  • 1 A $0.50\ \text{kg}$ block oscillates on a spring of constant $200\ \text{N/m}$. Find its period and frequency.
  • 2 A $0.25\ \text{kg}$ block on a spring makes $10$ oscillations in $6.28\ \text{s}$. Find the spring constant.
  • 3 What mass on a spring of constant $40\ \text{N/m}$ gives a period of $1.0\ \text{s}$?
  • 4 A spring–block oscillator is taken to the Moon. Does its period change? Explain.
  • 5 Adding $2.0\ \text{kg}$ to a block on a spring doubles the period. Find the original mass.
  • 6 A load hung from a spring stretches it by $4.9\ \text{cm}$. Find the period of small vertical oscillations of the load.
  • 7 A $1.0\ \text{kg}$ block on a spring of constant $100\ \text{N/m}$ is pulled $10\ \text{cm}$ from equilibrium and released. Find its maximum speed, maximum acceleration and total energy.
  • 8 A block on a spring has a period of $2.0\ \text{s}$; with an extra mass $\Delta m$ added, the period is $3.0\ \text{s}$. Find $\dfrac{\Delta m}{m}$.
1 $T=2\pi\sqrt{\dfrac{0.50}{200}}=0.31\ \text{s}$ ($0.314$); $\nu=3.2\ \text{Hz}$ ($3.18$).
2 $T=0.628\ \text{s}$; $k=\dfrac{4\pi^2m}{T^2}=\dfrac{39.48\times0.25}{0.394}=25\ \text{N/m}$.
3 $m=\dfrac{kT^2}{4\pi^2}=\dfrac{40}{39.48}=1.0\ \text{kg}$.
4 No: $T=2\pi\sqrt{m/k}$ contains neither $g$ nor anything else that changes. (A vertical spring hangs with a smaller stretch on the Moon, but $\sqrt{l_0/g}$ is still $\sqrt{m/k}$.)
5 $\sqrt{\dfrac{m+2.0}{m}}=2$: $m+2.0=4m$, $m=0.67\ \text{kg}$.
6 $T=2\pi\sqrt{\dfrac{l_0}{g}}=2\pi\sqrt{\dfrac{0.049}{9.8}}=0.44\ \text{s}$.
7 $\omega=10\ \text{rad/s}$: $v_\text{max}=1.0\ \text{m/s}$, $a_\text{max}=10\ \text{m/s}^2$, $E=\tfrac12(100)(0.10)^2=0.50\ \text{J}$.
8 $\dfrac{m+\Delta m}{m}=\left(\dfrac{3.0}{2.0}\right)^2=2.25$: $\dfrac{\Delta m}{m}=1.25$.
II
Energy, speed and the vertical spring.
  • 1 A $0.40\ \text{kg}$ block on a spring of constant $160\ \text{N/m}$ oscillates with amplitude $5.0\ \text{cm}$. Find $\omega$, the period, the maximum speed and the total energy.
  • 2 For the oscillator of part 1, find the speed and the kinetic energy when the displacement is $3.0\ \text{cm}$.
  • 3 At what displacement are the kinetic and potential energies of the oscillator in part 1 equal?
  • 4 A spring stretches by $2.5\ \text{cm}$ when a $0.50\ \text{kg}$ load hangs from it at rest. Find the spring constant and the frequency of small vertical oscillations of the load.
  • 5 Blocks of mass $m$ and $4m$ are attached in turn to the same spring. Find the ratio of their periods and of their frequencies.
  • 6 On one spring, a $0.10\ \text{kg}$ mass makes $25$ oscillations in $4.97\ \text{s}$ and a $0.40\ \text{kg}$ mass makes $25$ oscillations in $9.93\ \text{s}$. Show that $T^2\propto m$ and find the spring constant.
  • 7 Show that a block hanging from a spring experiences a net force $-ky$ when displaced by $y$ from its equilibrium position, and hence that $T=2\pi\sqrt{l_0/g}$, where $l_0$ is the static stretch.
1 $\omega=\sqrt{\dfrac{160}{0.40}}=20\ \text{rad/s}$; $T=\dfrac{2\pi}{20}=0.31\ \text{s}$; $v_\text{max}=\omega A=1.0\ \text{m/s}$; $E=\tfrac12kA^2=\tfrac12(160)(0.050)^2=0.20\ \text{J}$.
2 $v=\omega\sqrt{A^2-x^2}=20\sqrt{0.0025-0.0009}=20\times0.040=0.80\ \text{m/s}$; $K=\tfrac12(0.40)(0.80)^2=0.13\ \text{J}$ ($0.128$; check: $\tfrac12k(A^2-x^2)=0.128\ \text{J}$ ✓).
3 $\tfrac12kx^2=\tfrac12\cdot\tfrac12kA^2$: $x=\pm\dfrac{A}{\sqrt2}=\pm3.5\ \text{cm}$.
4 $k=\dfrac{0.50\times9.8}{0.025}=196\ \text{N/m}$ ($2.0\times10^2$); $\nu=\dfrac1{2\pi}\sqrt{\dfrac{g}{l_0}}=\dfrac1{2\pi}\sqrt{\dfrac{9.8}{0.025}}=3.2\ \text{Hz}$.
5 $T\propto\sqrt m$: $T_1:T_2=1:2$; frequencies $2:1$.
6 $T=0.1988\ \text{s}$ and $0.3972\ \text{s}$; $T^2=0.03952$ and $0.1578\ \text{s}^2$, ratio $3.99\approx4$ for a mass ratio of $4$ ✓. $k=\dfrac{4\pi^2m}{T^2}=\dfrac{39.48\times0.10}{0.03952}=1.0\times10^2\ \text{N/m}$ (the second mass gives the same).
7 At equilibrium $kl_0=mg$. Displaced by $y$ downward, the spring is stretched by $l_0+y$: net downward force $mg-k(l_0+y)=-ky$ — a linear restoring force, so SHM with $\omega^2=\dfrac km=\dfrac g{l_0}$ and $T=2\pi\sqrt{\dfrac{l_0}{g}}$.
III
The simple pendulum.
  • 1 Find the period of a simple pendulum $2.45\ \text{m}$ long.
  • 2 What length of simple pendulum has a period of $1.5\ \text{s}$?
  • 3 The length of a pendulum is increased by $21\%$. By what percentage does its period increase?
  • 4 Find the period of a $0.80\ \text{m}$ pendulum on Mars, where $g=3.7\ \text{m/s}^2$.
  • 5 Find the length of a seconds pendulum on the Moon ($g=1.62\ \text{m/s}^2$).
  • 6 Explain why the period of a simple pendulum does not depend on the mass of the bob.
  • 7 A pendulum $1.0\ \text{m}$ long is released from $8.0^\circ$. Find the bob's speed at the lowest point (a) from energy conservation, (b) from $v_\text{max}=\omega A$ with $A=l\theta_0$.
  • 8 In a laboratory, $20$ oscillations of a pendulum $1.000\ \text{m}$ long take $40.1\ \text{s}$. Find $g$.
1 $T=2\pi\sqrt{\dfrac{2.45}{9.8}}=2\pi\times0.50=3.1\ \text{s}$ ($3.14$).
2 $l=\dfrac{gT^2}{4\pi^2}=\dfrac{9.8\times2.25}{39.48}=0.56\ \text{m}$.
3 $T\propto\sqrt l$: $\sqrt{1.21}=1.10$, a $10\%$ increase.
4 $T=2\pi\sqrt{\dfrac{0.80}{3.7}}=2.9\ \text{s}$ ($2.92$).
5 $l=\dfrac{g}{\pi^2}=\dfrac{1.62}{9.87}=0.164\ \text{m}$.
6 The restoring torque $mgl\sin\theta$ and the moment of inertia $ml^2$ are both proportional to $m$, so $m$ cancels from $\alpha=-\dfrac gl\sin\theta$ — gravitational mass and inertial mass are the same.
7 (a) $v=\sqrt{2gl(1-\cos8.0^\circ)}=\sqrt{19.6\times0.00973}=0.44\ \text{m/s}$. (b) $\omega=\sqrt{9.8}=3.13\ \text{rad/s}$, $A=1.0\times0.1396=0.140\ \text{m}$: $v_\text{max}=0.44\ \text{m/s}$ — the same to two figures at this small angle.
8 $T=2.005\ \text{s}$; $g=\dfrac{4\pi^2l}{T^2}=\dfrac{39.48}{4.020}=9.82\ \text{m/s}^2$.
IV
Applications and harder problems.
  • 1 (Modelling) A pendulum clock that is correct at sea level is taken to a place $3.2\ \text{km}$ higher. How many seconds per day does it lose? ($R_E=6.37\times10^6\ \text{m}$.)
  • 2 (Modelling) When four passengers of total mass $300\ \text{kg}$ climb into a $1200\ \text{kg}$ car, its body sinks $3.0\ \text{cm}$ on its springs. Find the effective spring constant of the suspension and the period and frequency with which the loaded car bounces.
  • 3 A pendulum $1.0\ \text{m}$ long hangs in a lift. Find its period when the lift (a) is at rest; (b) accelerates upward at $2.2\ \text{m/s}^2$; (c) falls freely. (Hint: in the lift's frame the bob behaves as if $g$ were replaced by the apparent $g$.)
  • 4 Explain why, when measuring $g$ with a pendulum, the amplitude must be kept small, and why it helps to time many oscillations and to use several lengths.
  • 5 A $0.20\ \text{kg}$ block hangs from a spring of constant $20\ \text{N/m}$. It is pulled down $5.0\ \text{cm}$ below its equilibrium position and released. Find the period. Does the spring ever become slack? What is the largest amplitude for which it does not?
  • 6 Pendulums $1.00\ \text{m}$ and $1.21\ \text{m}$ long are released together, in phase. After how many oscillations of the longer pendulum are they next in phase, and how long does this take?
  • 7 A spring–block oscillator and a simple pendulum both have period $0.50\ \text{s}$ on the Earth. Find their periods on the Moon ($g=1.62\ \text{m/s}^2$).
1 $\dfrac{T_h}{T}=1+\dfrac{h}{R_E}$: fraction $\dfrac{3200}{6.37\times10^6}=5.02\times10^{-4}$; loss $86\,400\times5.02\times10^{-4}=43\ \text{s}$ per day.
2 $k=\dfrac{300\times9.8}{0.030}=9.8\times10^4\ \text{N/m}$. $T=2\pi\sqrt{\dfrac{1500}{9.8\times10^4}}=0.78\ \text{s}$; $\nu=1.3\ \text{Hz}$ ($1.29$).
3 (a) $2\pi\sqrt{1.0/9.8}=2.0\ \text{s}$ ($2.01$). (b) Apparent $g=9.8+2.2=12.0\ \text{m/s}^2$: $T=2\pi\sqrt{1.0/12.0}=1.8\ \text{s}$ ($1.81$). (c) Apparent $g=0$: there is no restoring torque, and the pendulum does not oscillate at all.
4 $T=2\pi\sqrt{l/g}$ assumes $\sin\theta\approx\theta$; at large amplitudes the true period is longer and the calculated $g$ too small. Timing $20$ or more oscillations divides the reaction-time error by $20$; several lengths and a $T^2$–$l$ graph average out random errors and expose a systematic error in $l$ (a line missing the origin).
5 $T=2\pi\sqrt{\dfrac{0.20}{20}}=0.63\ \text{s}$. Equilibrium stretch $l_0=\dfrac{0.20\times9.8}{20}=9.8\ \text{cm}$, greater than $A=5.0\ \text{cm}$, so the spring stays stretched throughout. It would go slack for $A\gt9.8\ \text{cm}$.
6 $\dfrac{T_2}{T_1}=\sqrt{1.21}=1.1$. They are next in phase when the shorter has made one more oscillation: $1.1n=n+1$, $n=10$ oscillations of the longer, taking $10\times2\pi\sqrt{1.21/9.8}=22\ \text{s}$.
7 Spring: $0.50\ \text{s}$ (unchanged). Pendulum: $0.50\sqrt{\dfrac{9.8}{1.62}}=1.2\ \text{s}$ ($1.23$).
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks Show that, for small amplitudes, the motion of a simple pendulum is simple harmonic, and derive the expression for its period. State the approximation you use.

Torque Bob of mass $m$ on a string of length $l$, at angle $\theta$. About the support, the tension has no torque; the weight gives $\tau=-mgl\sin\theta$ (restoring). [1]
Equation $I=ml^2$ and $\tau=I\alpha$: $\alpha=-\dfrac gl\sin\theta$. For small $\theta$ (in radians) $\sin\theta\approx\theta$, so $\alpha=-\dfrac gl\theta$: angular acceleration proportional to, and opposite to, angular displacement — SHM. [1]
Period Comparing with $\alpha=-\omega^2\theta$: $\omega=\sqrt{\dfrac gl}$, $T=2\pi\sqrt{\dfrac lg}$. [1]
Exam tip Draw the bob with $mg$ resolved into $mg\cos\theta$ (along the string) and $mg\sin\theta$ (along the arc), and write "valid for $\theta\lesssim10^\circ$".

E2. NEET A block of mass $m$ is attached to two identical springs, each of spring constant $k$, joined end to end (in series). The period of oscillation is (a) $2\pi\sqrt{\dfrac{m}{2k}}$ (b) $2\pi\sqrt{\dfrac mk}$ (c) $2\pi\sqrt{\dfrac{2m}{k}}$ (d) $4\pi\sqrt{\dfrac mk}$

Note Springs in combination are beyond the NCERT/CBSE syllabus (see the 🚀 box); this is entrance-exam practice.
Idea In series $\dfrac1{k_\text{s}}=\dfrac1k+\dfrac1k$, so $k_\text{s}=\dfrac k2$.
Answer $T=2\pi\sqrt{\dfrac{m}{k/2}}=2\pi\sqrt{\dfrac{2m}{k}}$: (c).
Exam tip (a) is the parallel answer. Series is softer, so the period must be longer than with one spring — a quick check that rules out (a) and (b).

E3. JEE Main A block on a spring oscillates with period $T$. The spring is cut into two equal halves and the same block is attached to one half. The new period is (a) $\dfrac T2$ (b) $\dfrac{T}{\sqrt2}$ (c) $\sqrt2\,T$ (d) $2T$

Note Cutting a spring is beyond the NCERT/CBSE syllabus (see the 🚀 box); this is entrance-exam practice.
Idea $k\propto\dfrac1{\text{length}}$: each half has constant $2k$.
Answer $T'=2\pi\sqrt{\dfrac m{2k}}=\dfrac{T}{\sqrt2}$: (b).
Exam tip (a) uses $T\propto\dfrac1k$ instead of $\dfrac1{\sqrt k}$. If both halves were used in parallel, $k_\text{eff}=4k$ and the period would be $\dfrac T2$.

E4. JEE Main (Numerical value) A block hanging at rest from a light spring stretches it by $6.25\ \text{cm}$. Find the frequency, in hertz, of its small vertical oscillations. Take $g=\pi^2\ \text{m/s}^2$.

Idea $kl_0=mg$, so $\omega=\sqrt{\dfrac km}=\sqrt{\dfrac g{l_0}}$.
Numbers $\omega=\sqrt{\dfrac{\pi^2}{0.0625}}=4\pi\ \text{rad/s}$; $\nu=\dfrac{\omega}{2\pi}=2$.
Answer $2$ (Hz).
Exam tip Neither $m$ nor $k$ is needed — only their ratio, which the static stretch gives as $\dfrac{l_0}{g}$.

E5. JEE Advanced (One or more correct) Which statements are correct? (A) The period of a simple pendulum swinging through small angles does not depend on the mass of the bob. (B) The period of a block on a vertical spring does not depend on $g$. (C) The period of a simple pendulum in a lift decreases when the lift accelerates upward. (D) Doubling the amplitude of a block on a spring (within Hooke's law) doubles its period.

(A) $T=2\pi\sqrt{l/g}$ has no $m$ ✓.
(B) Gravity shifts the equilibrium by $\dfrac{mg}k$ but $T=2\pi\sqrt{m/k}$ ✓.
(C) The apparent $g$ is $g+a$, larger, so $T=2\pi\sqrt{\dfrac{l}{g+a}}$ is shorter ✓.
(D) SHM's period is independent of amplitude ✗; the maximum speed doubles instead.
Answer (A), (B), (C).
Exam tip For lift questions, replace $g$ by the apparent $g$ for the pendulum — and leave the spring–block period alone.

E6. Board · Case study In a school laboratory, students time $20$ oscillations of a simple pendulum for several lengths and plot $T^2$ against $l$. Their best straight line passes through the origin with a slope of $4.0\ \text{s}^2/\text{m}$. (i) Why do they plot $T^2$ rather than $T$ against $l$? [1] (ii) Find the value of $g$ they obtain. [2] (iii) Give one reason why the bob should swing through only a small angle. [1]

(i) $T^2=\dfrac{4\pi^2}{g}l$ is a straight line through the origin, whose slope is easy to measure; $T$ against $l$ is a curve. [1]
(ii) $g=\dfrac{4\pi^2}{\text{slope}}=\dfrac{39.48}{4.0}=9.9\ \text{m/s}^2$ ($9.87$). [2]
(iii) The formula assumes $\sin\theta\approx\theta$; at large angles the period is longer than $2\pi\sqrt{l/g}$, so $g$ would come out too small. [1]
Exam tip Mention that $l$ is measured to the centre of the bob — a common one-mark question.
🔬 Research–Teaching NexusResearch-tutored · optional, about 60 minutes

Research question. How accurately can you measure the local value of $g$ with a pendulum made from string and a nut — and can you detect that $g$ is different from the "standard" $9.8\ \text{m/s}^2$?

Why it matters. Until the twentieth century, precise pendulum timings were the main way of measuring $g$ around the world, and they revealed that the Earth is flattened at the poles. Geophysicists still map small variations of $g$ (with modern gravimeters) to locate ore bodies, oil-bearing rocks and underground cavities, and to monitor groundwater and volcanoes. National measurement institutes publish an accurate local $g$ for every laboratory that calibrates weighing instruments.

What to do. Hang a heavy nut or small metal ball from a thin, inextensible thread clamped firmly between two coins or blocks.

  1. Measure the length from the clamp to the centre of the bob. Start with about $1.0\ \text{m}$.
  2. Pull the bob aside by about $5^\circ$ (a few centimetres) and release it. Time $20$ oscillations, starting the count as the bob passes the lowest point. Repeat three times.
  3. Repeat for at least five lengths between $0.3\ \text{m}$ and $1.5\ \text{m}$.
  4. Plot $T^2$ against $l$, draw the best straight line, and find $g$ from its slope. Does the line pass through the origin? What would an offset mean?
  5. Repeat one length with an amplitude of $30^\circ$ and $45^\circ$. Can you detect the longer period?

Tools: Thread, clamp, metre rule, protractor, phone stopwatch (or slow-motion video with a timer in view), spreadsheet or graph paper.

Hand in: Your data table with repeat timings; the $T^2$–$l$ graph; your value of $g$ with an uncertainty estimate (from the spread of the timings); a comparison with the expected local value; and a list of the main systematic errors.

Going further. Look up the formula for $g$ at sea level as a function of latitude, and the correction for height above sea level. Is your measurement precise enough to tell your city from one at the equator?

Instructor note: sea-level $g$ varies from about $9.780\ \text{m/s}^2$ at the equator to $9.832\ \text{m/s}^2$ at the poles; a common approximation is $g\approx9.780\,(1+0.0053\sin^2\varphi)$ at latitude $\varphi$, minus about $3.1\times10^{-6}\ \text{m/s}^2$ per metre of height. For Delhi ($\varphi\approx28.6^\circ$, about $220\ \text{m}$) this gives $9.792$ at sea level and about $9.791\ \text{m/s}^2$ after the small height correction ($-0.0007$). Careful students reach $\pm0.5\%$ ($\pm0.05\ \text{m/s}^2$), which is not enough to see the latitude effect (about $0.5\%$ pole to equator) — a useful discussion of precision. At $30^\circ$ the period is $1.7\%$ longer; at $45^\circ$ about $4\%$, which good timings can detect.

🧪 Self-check quiz

0/5 correct

Q1. The mass of a block on a spring is made four times as large. Its period

Q2. The period of a simple pendulum swinging through a small angle depends on

Q3. On the Earth ($g=9.8\ \text{m/s}^2$) a seconds pendulum has a length of about

Q4. A block hanging at rest stretches a spring by $9.8\ \text{cm}$. The period of its small vertical oscillations is about

Q5. A spring–block oscillator and a simple pendulum are taken from the Earth to the Moon. Then

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