Simple Harmonic Motion — Springs, Pendulums, and Beyond

Push a swing, stretch a spring, or pull a pendulum — let go, and the object oscillates back and forth in a predictable, beautiful pattern. This is Simple Harmonic Motion (SHM), and it’s one of the most important concepts in physics.

SHM appears everywhere: the vibration of atoms in a crystal lattice, the oscillation of AC current, the vibration of molecules (responsible for heat), the behavior of LC circuits in electronics, and even the vibrations of the human eardrum. Every student who wants to do well in JEE Main or Class 12 boards must build a solid intuition for SHM — it’s not enough to memorize formulas.

Key Terms and Definitions

Oscillation: A back-and-forth motion about an equilibrium position.

Simple Harmonic Motion: A specific type of oscillation where the restoring force is directly proportional to displacement and directed towards the equilibrium position. Mathematically: $F = -kx$ .

Amplitude (A): The maximum displacement from the equilibrium position. Unit: metres.

Time period (T): The time taken to complete one full oscillation. Unit: seconds.

Frequency (f): Number of complete oscillations per second. $f = 1/T$ . Unit: Hertz (Hz).

Angular frequency (ω): $\omega = 2\pi f = 2\pi/T$ . Unit: rad/s. This appears in all SHM equations.

Phase: Describes the position and direction of motion of the particle at any instant. The phase of $x = A\sin(\omega t + \phi)$ at time $t$ is $(\omega t + \phi)$ .

Equilibrium position: The position where the net force on the particle is zero. The particle returns to this position after each displacement.

Restoring force: The force that brings the displaced particle back toward equilibrium. Always directed toward equilibrium, hence the negative sign in $F = -kx$ .

The Mathematical Description of SHM

Equation of motion

From Newton’s second law: $ma = -kx$ , which gives:

\frac{d^2x}{dt^2} = -\omega^2 x \quad \text{where } \omega^2 = k/m

This second-order differential equation has the general solution:

x(t) = A\sin(\omega t + \phi)

where $A$ is the amplitude, $\omega$ is angular frequency, and $\phi$ is the initial phase (depends on initial conditions).

Velocity and acceleration in SHM

Differentiating $x = A\sin(\omega t + \phi)$ :

v = \frac{dx}{dt} = A\omega\cos(\omega t + \phi)

a = \frac{dv}{dt} = -A\omega^2\sin(\omega t + \phi) = -\omega^2 x

Key insight: velocity leads displacement by 90° (quarter cycle), and acceleration leads velocity by 90° (acceleration is exactly opposite to displacement).

Displacement: $x = A\sin(\omega t + \phi)$

Velocity: $v = A\omega\cos(\omega t + \phi)$

Also: $v = \omega\sqrt{A^2 - x^2}$ ← useful when you know x, not t

Acceleration: $a = -\omega^2 x$

Velocity is maximum (= $A\omega$ ) at equilibrium ( $x = 0$ )

Velocity is zero (= 0) at extremes ( $x = \pm A$ )

Acceleration is maximum ( $= A\omega^2$ ) at extremes ( $x = \pm A$ )

Acceleration is zero at equilibrium ( $x = 0$ )

Energy in SHM

The energy in SHM alternates between kinetic and potential, but the total mechanical energy remains constant.

E_{kinetic} = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2(A^2 - x^2)

E_{potential} = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2x^2

E_{total} = E_K + E_P = \frac{1}{2}m\omega^2A^2 = \frac{1}{2}kA^2

The total energy depends on $A^2$ — doubling the amplitude quadruples the energy.

$E_{total} = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2A^2$

KE is maximum at equilibrium; PE is maximum at extremes.

At $x = A/\sqrt{2}$ : KE = PE = $E_{total}/2$ (equal energy partition)

Common SHM Systems

Spring-mass system

A mass $m$ attached to a spring with spring constant $k$ :

T = 2\pi\sqrt{\frac{m}{k}}, \quad \omega = \sqrt{\frac{k}{m}}

The period is independent of amplitude (for ideal springs) — this is a crucial property of SHM.

Two springs in series (effective spring constant): $\frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2}$

Two springs in parallel (effective spring constant): $k_{eff} = k_1 + k_2$

Simple pendulum

A point mass on a string of length $L$ oscillating with small angles ( $\theta < 15°$ ):

T = 2\pi\sqrt{\frac{L}{g}}, \quad \omega = \sqrt{\frac{g}{L}}

The period is independent of mass and amplitude (for small angles) — only the length and $g$ matter.

A common JEE trick: if $g$ changes (pendulum in an elevator, at a different height, etc.), the pendulum period changes too. An accelerating elevator effectively changes $g$ : $g_{eff} = g \pm a$ . Acceleration upward → $g_{eff}$ increases → period decreases → pendulum runs fast. This appears frequently in JEE Main problems about pendulums in non-inertial frames.

Liquid in a U-tube

A liquid of density $\rho$ oscillating in a U-tube of length $L$ (total liquid column):

T = 2\pi\sqrt{\frac{L}{2g}}

Solved Examples

Example 1 (CBSE Level)

Q: A particle in SHM has amplitude 10 cm and period 2 s. Find (a) angular frequency, (b) maximum velocity, (c) maximum acceleration.

Solution: (a) $\omega = 2\pi/T = 2\pi/2 = \pi$ rad/s

(b) $v_{max} = A\omega = 0.1 \times \pi = 0.314$ m/s (at equilibrium)

Example 2 (JEE Main Level)

Q: A spring of spring constant 200 N/m is stretched by 5 cm and a mass of 0.5 kg is attached. If released from rest, find the time period and total energy of oscillation.

Solution: Amplitude $A = 5$ cm = 0.05 m (since released from rest at maximum displacement)

$T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.5/200} = 2\pi\sqrt{0.0025} = 2\pi \times 0.05 = 0.314$ s

$E = \frac{1}{2}kA^2 = \frac{1}{2} \times 200 \times (0.05)^2 = \frac{1}{2} \times 200 \times 0.0025 = 0.25$ J

Example 3 (JEE Advanced Level)

Q: Two particles undergo SHM along the same line with the same amplitude $A$ and time period $T$ , but differ in phase by $\pi/3$ . Find the separation between the particles when their velocity vectors are in opposite directions.

Solution: Let $x_1 = A\sin\omega t$ and $x_2 = A\sin(\omega t + \pi/3)$ .

Velocities are in opposite directions when their velocities have opposite signs — this requires the particles to be moving in opposite directions.

This is a more involved problem. The maximum separation occurs when the phase difference is $\pi/3$ . Using the formula for separation between two SHM particles with phase difference $\delta$ :

$x_2 - x_1 = A\sin(\omega t + \pi/3) - A\sin\omega t = 2A\cos(\omega t + \pi/6)\sin(\pi/6)$

$= 2A \times \frac{1}{2} \times \cos(\omega t + \pi/6) = A\cos(\omega t + \pi/6)$

Maximum separation = $A$ . When velocities are opposite (one going right, one going left), the phase difference condition gives separation = $A$ .

Exam-Specific Tips

JEE Main SHM Weightage: SHM consistently appears with 2-3 questions per shift. Most common question types: (1) velocity/acceleration at a given displacement, (2) period of compound systems (springs in series/parallel), (3) energy distribution, (4) pendulum in elevators or in liquids. The formula $v = \omega\sqrt{A^2 - x^2}$ appears in almost every SHM problem.

CBSE Class 11 Board: SHM is in Chapter 14 (Oscillations). Long answer questions (5 marks) typically ask for derivation of velocity in SHM or derivation of period of a spring-mass system. Write $F = -kx$ , apply Newton’s second law, arrive at the differential equation, then state the solution.

Quick verification: For any SHM problem, always check that $a = -\omega^2 x$ holds. If your $a$ and $x$ have the same sign (both positive or both negative), something is wrong — the restoring force must oppose displacement.

Common Mistakes to Avoid

Mistake 1: Using $T = 2\pi\sqrt{m/k}$ for a pendulum. The pendulum formula is $T = 2\pi\sqrt{L/g}$ . The spring formula uses $k$ (spring constant) and $m$ (mass). These are two different systems with different parameters.

Mistake 2: Thinking maximum velocity occurs at the extremes. Maximum velocity occurs at the equilibrium position (where KE is maximum and PE is zero). At extremes, velocity = 0 (the particle momentarily stops before reversing). This is a classic confusion.

Mistake 3: Forgetting that $\omega^2 = k/m$ , not $\omega = k/m$ . When finding $\omega$ from the equation $F = -kx$ : $ma = -kx \Rightarrow a = -(k/m)x$ . Comparing with $a = -\omega^2 x$ gives $\omega^2 = k/m$ . Take the square root to get $\omega$ .

Mistake 4: Not checking the small angle condition for pendulums. The pendulum formula $T = 2\pi\sqrt{L/g}$ is valid only for small amplitudes (typically $\theta < 15°$ ). For large amplitude oscillations, the period is longer and the formula doesn’t apply exactly. JEE sometimes tests whether you know this limitation.

Mistake 5: Confusing phase with initial phase. The phase at time $t$ is $(\omega t + \phi)$ ; the initial phase is just $\phi$ (at $t = 0$ ). When a problem says “starts from equilibrium with velocity in positive direction,” the initial conditions give $x(0) = 0$ and $v(0) > 0$ , which means $\phi = 0$ and the equation is $x = A\sin\omega t$ .

Practice Questions

Q1: A particle in SHM has position $x = 5\sin(3t + \pi/4)$ cm. Find the amplitude, angular frequency, time period, and position at $t = 0$ .

Comparing with $x = A\sin(\omega t + \phi)$ :

Amplitude $A = 5$ cm

Angular frequency $\omega = 3$ rad/s

Time period $T = 2\pi/\omega = 2\pi/3 \approx 2.09$ s

At $t = 0$ : $x = 5\sin(\pi/4) = 5 \times (1/\sqrt{2}) = 5/\sqrt{2} \approx 3.54$ cm

Q2: A spring-mass system has period 1 s. If the spring constant is doubled and mass is halved, what is the new period?

Original: $T_1 = 2\pi\sqrt{m/k} = 1$ s

New: $T_2 = 2\pi\sqrt{(m/2)/(2k)} = 2\pi\sqrt{m/(4k)} = \frac{1}{2} \times 2\pi\sqrt{m/k} = T_1/2 = 0.5$ s

The period halves.

Q3: At what displacement is the kinetic energy equal to potential energy in SHM?

KE = PE when $\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2$

$A^2 - x^2 = x^2 \Rightarrow A^2 = 2x^2 \Rightarrow x = A/\sqrt{2} = 0.707A$

At $x = A/\sqrt{2}$ (approximately 71% of amplitude), KE = PE = total energy/2.

Q4: A pendulum of length 1 m swings on the surface of the moon where $g = 1.6\text{ m/s}^2$ . What is its period?

$T = 2\pi\sqrt{L/g} = 2\pi\sqrt{1/1.6} = 2\pi\sqrt{0.625} = 2\pi \times 0.79 \approx 4.97$ s

Compare: on Earth with $g = 9.8$ , $T = 2\pi\sqrt{1/9.8} \approx 2.0$ s. The lunar pendulum is much slower because gravity is weaker.

FAQs

Q: Why is SHM called “simple”? The word “simple” refers to the simplicity of the force law ( $F = -kx$ — linear in displacement). More complex oscillations involve restoring forces that are nonlinear (e.g., $F = -kx^3$ ). A simple pendulum at small angles is approximately SHM because $\sin\theta \approx \theta$ for small $\theta$ , making the restoring force approximately linear.

Q: Is all oscillatory motion SHM? No — SHM is a special case. All SHM is oscillatory, but not all oscillatory motion is SHM. A ball bouncing on the floor oscillates, but the restoring force isn’t proportional to displacement. A pendulum at large angles is oscillatory but not SHM (force is proportional to $\sin\theta$ , not $\theta$ ). The defining condition for SHM is $F \propto -x$ .

Q: Why is the period of SHM independent of amplitude? This is the isochronous property of SHM. Mathematically, $T = 2\pi/\omega = 2\pi\sqrt{m/k}$ — there’s no $A$ in the formula. Physically: a larger amplitude means greater restoring force (since $F = -kx$ ), which gives greater acceleration, which compensates for the longer distance traveled. The particle “speeds up” proportionally to how far it needs to go.

Q: How does damping affect SHM? Real-world oscillations are damped — energy is lost to friction or air resistance. In damped SHM, the amplitude decreases exponentially over time while the period remains approximately the same (for light damping). Critically damped systems return to equilibrium without oscillating. Overdamped systems return even more slowly. This is Class 12 content in CBSE and JEE.