Simple harmonic motion for JEE Main

Simple harmonic motion looks like a list of formulas, but it is really one test: is the restoring force proportional to the displacement? If F = −kx (or τ = −cθ), the motion is SHM, and ω = √(k/m) follows at once. Everything else in the chapter, from energy to phase to pendulums in lifts, builds on that.

The syllabus covers periodic motion, SHM and its equation, phase, oscillations of a spring, energy in SHM and the simple pendulum with a derivation of its period. SHM also returns in waves and alternating current, so time spent here pays off later.

What makes motion simple harmonic

Periodic motion repeats; oscillatory motion goes back and forth about a mean position. SHM is the special oscillation in which:

F = −kx, so a = −ω²x, with ω = √(k/m)

The minus sign means the force always points back towards the mean position. Period T = 2π/ω and frequency f = 1/T.

To identify SHM in a new situation, displace the body by a small x, find the net restoring force, and write it as −(constant) × x. That constant is the effective k.

Displacement, velocity and acceleration

With amplitude A and initial phase φ:

• x = A sin(ωt + φ)
• v = Aω cos(ωt + φ) = ±ω√(A² − x²)
• a = −ω²x

Velocity leads displacement by π/2, and acceleration leads it by π (they are opposite). A v–x graph is an ellipse; an a–x graph is a straight line through the origin with negative slope.

Using phase to find times. Measure time from the mean position, so x = A sin ωt. The body reaches x = A/2 when ωt = π/6, that is at t = T/12. It then takes a further T/4 − T/12 = T/6 to reach the extreme. Half the distance takes one-third of the time of the other half, because the body is fastest near the mean.

Energy in SHM

• Potential energy: U = ½kx² = ½mω²x²
• Kinetic energy: K = ½mω²(A² − x²)
• Total: E = ½kA² = ½mω²A², constant

KE equals PE when x = A/√2. Averaged over a cycle, KE and PE are each E/2. Both KE and PE oscillate at twice the frequency of the motion.

Spring–mass systems

T = 2π√(m/k). The period does not depend on g, so a mass on a vertical spring has the same period as on a horizontal one. The only change is the equilibrium position, which is lower by mg/k. That gives a quick shortcut: T = 2π√(Δ/g), where Δ is the static extension.

The simple pendulum

A bob of mass m hangs on a string of length L. Displaced by a small angle θ, gravity provides a restoring torque −mgL sin θ ≈ −mgLθ. With I = mL²:

α = −(g/L)θ, so T = 2π√(L/g).

This needs small angles (sin θ ≈ θ) and a light string. The period does not depend on the mass or the amplitude.

A seconds pendulum has T = 2 s; its length is close to 1 m. When the pendulum's support accelerates, replace g by the effective g:

Worked problems

Problem 1: speed at a point. A particle in SHM has A = 0.2 m and ω = 5 rad/s. Find its speed at x = 0.12 m.

v = ω√(A² − x²) = 5√(0.04 − 0.0144) = 5 × 0.16 = 0.8 m/s.

Problem 2: reading an equation. x = 5 sin(πt + π/3), with x in cm and t in s. Find the period, maximum speed and maximum acceleration.

ω = π rad/s, so T = 2 s.
v_max = Aω = 5π ≈ 15.7 cm/s.
a_max = ω²A = 5π² ≈ 49.3 cm/s².

Problem 3 (numerical answer): spring energy. A 1 kg block on a spring of k = 400 N/m is pulled 5 cm from equilibrium and released on a smooth floor. Find the period, the energy and the maximum speed.

ω = √(400/1) = 20 rad/s, so T = 2π/20 = π/10 ≈ 0.314 s.
E = ½ × 400 × 0.05² = 0.5 J.
v_max = Aω = 0.05 × 20 = 1 m/s. Check: ½ × 1 × 1² = 0.5 J.

Problem 4: pendulum in a lift. A pendulum has a period of 2 s in a stationary lift. Find its period when the lift accelerates upwards at g/3.

T′ = T√(g/g_eff) = 2√(g/(4g/3)) = 2√(3/4) = √3 ≈ 1.73 s.

Practice set

1. At what displacement is the kinetic energy equal to the potential energy?
2. At what displacement is the kinetic energy three times the potential energy?
3. A spring of constant k is cut into two equal halves. Find the constant of each half, and of the two halves joined in parallel.
4. A mass m hangs from two identical springs (constant k each). Compare the periods when the springs are in series and in parallel.
5. A particle in SHM has a period of 12 s. Find the minimum time to go from the mean position to half the amplitude, and from there to the extreme.
6. A 2 kg mass stretches a spring by 5 cm when it hangs at rest (g = 10 m/s²). Find k and the period of vertical oscillations.
7. A particle has amplitude 4 cm and maximum speed 8 cm/s. Find ω and the maximum acceleration.
8. A simple pendulum has a period of 2 s on Earth. What is its period on a planet where g is one-quarter of Earth's?

Answers

1. A/√2.
2. K = 3U means U = E/4, so x² = A²/4 and x = A/2.
3. Each half: 2k. Both in parallel: 4k.
4. Series: k/2, so T = 2π√(2m/k). Parallel: 2k, so T = 2π√(m/2k). Ratio 2 : 1.
5. T/12 = 1 s; then T/6 = 2 s.
6. k = 20/0.05 = 400 N/m. T = 2π√(2/400) = 2π√(0.05/10) ≈ 0.44 s.
7. ω = 8/4 = 2 rad/s; a_max = ω²A = 16 cm/s².
8. T ∝ 1/√g, so T = 2 × 2 = 4 s.

What to do next

• Derive T = 2π√(L/g) and T = 2π√(m/k) from the restoring force.
• Practise phase questions (time from x₁ to x₂) until you can do them without a diagram.
• Solve 20 problems on spring combinations and pendulums with an effective g.
• Move on to waves and sound, and revise rotational dynamics for the torque form of the pendulum derivation.