1. Physics
  2. Oscillations and Waves
  3. Simple Harmonic Motion (SHM)

Oscillations and Waves · JEE & NEET Physics

Simple Harmonic Motion (SHM): notes and previous year questions

The restoring force F = −kx, x = A sin(ωt + φ), velocity and acceleration, period, springs, pendulums and phase.

Simple Harmonic Motion (SHM) in short

  • SHM is periodic, oscillatory motion with a restoring force proportional to the displacement: F = −kx.
  • The shadow of a point moving round a circle does SHM, so x = A sin(ωt + φ).
  • A is the amplitude, ω the angular frequency, ωt + φ the phase and φ the phase at t = 0.
  • The speed is largest, Aω, at the centre; the acceleration is largest, Aω², at the ends.

1What is simple harmonic motion?

A swinging pendulum or a block bouncing on a spring repeats its motion after a fixed time, so it is periodic, and it goes to and fro about a centre, the equilibrium position, so it is oscillatory. The simplest oscillation of all is simple harmonic motion (SHM).

Pull a block on a spring to the right and the spring pulls it back to the left; push it left and the spring pushes it right. The farther it goes, the stronger the pull: twice the displacement, twice the force. This force, always pointing back to the centre, is the restoring force.

F=−kxF = -kxThe definition of SHM: restoring force proportional to the displacement x and opposite to it.
  • kk is the force constant (N/m): how stiff the spring is.
  • The minus sign means the force always points back toward x=0x = 0.
  • Example: k=100k = 100 N/m and x=+0.20x = +0.20 m give F=−20F = -20 N, 20 N to the left.

2The SHM equation

A point going round a circle of radius AA at a steady ω\omega radians per second casts a shadow on a diameter. The shadow moves fastest in the middle and stops at the ends: it does SHM. Plotted against time, its position is a sine curve.

x=Asin⁡(ωt+ϕ)x = A\sin(\omega t + \phi)
  • AA: the amplitude, the largest displacement (m).
  • ω\omega: the angular frequency (rad/s), how fast the point goes round the circle.
  • ωt+ϕ\omega t + \phi: the phase; ϕ\phi is the phase at t=0t = 0 (the phase constant).
  • x=Acos⁡(ωt+ϕ′)x = A\cos(\omega t + \phi') is the same motion with ϕ′=ϕ−π/2\phi' = \phi - \pi/2.
  • To start at x=+Ax = +A we need sin⁡ϕ=1\sin\phi = 1, so ϕ=π/2\phi = \pi/2 and x=Acos⁡ωtx = A\cos\omega t.

3Velocity and acceleration

At the centre the block moves fastest but the spring is relaxed, so the acceleration is zero. At the ends it stops for an instant but the spring pulls hardest, so the acceleration is largest.

v=dxdt=Aωcos⁡(ωt+ϕ)v = \frac{dx}{dt} = A\omega\cos(\omega t + \phi)
v=ωA2−x2v = \omega\sqrt{A^2 - x^2}Using cos² + sin² = 1 to write v in terms of x.
a=dvdt=−ω2xa = \frac{dv}{dt} = -\omega^2 xThe test for SHM: acceleration proportional to displacement and opposite to it.
QuantityFormulaLargest valueWhere
Displacementx=Asin⁡(ωt+ϕ)x = A\sin(\omega t + \phi)AAthe ends
Speedv=ωA2−x2v = \omega\sqrt{A^2 - x^2}vmax=Aωv_{max} = A\omegathe centre, x=0x = 0
Accelerationa=−ω2xa = -\omega^2 xamax=Aω2a_{max} = A\omega^2the ends, x=±Ax = \pm A

4Period and frequency

One full oscillation is a round trip: centre → one end → centre → other end → centre. Its time is the period TT. The number of oscillations per second is the frequency ff, in hertz (1 Hz = 1 s⁻¹). One oscillation is one turn of the circle, 2π2\pi radians.

T=2πω,f=1T=ω2πT = \frac{2\pi}{\omega}, \quad f = \frac{1}{T} = \frac{\omega}{2\pi}
ω=2πf\omega = 2\pi f

5Spring and mass

The spring gives F=−kxF = -kx and Newton's second law gives F=maF = ma, so a=−(k/m)xa = -(k/m)x. Comparing with a=−ω2xa = -\omega^2 x:

ω=km,T=2πmk\omega = \sqrt{\frac{k}{m}}, \quad T = 2\pi\sqrt{\frac{m}{k}}
f=12πkmf = \frac{1}{2\pi}\sqrt{\frac{k}{m}}
  • A 4 times heavier block swings twice as slowly (T∝mT \propto \sqrt{m}); a stiffer spring swings faster (T∝1/kT \propto 1/\sqrt{k}).
  • The period does not depend on the amplitude, and a spring's period does not depend on g.
  • k=200k = 200 N/m and m=2m = 2 kg: T=2π0.01=0.2π≈0.628T = 2\pi\sqrt{0.01} = 0.2\pi \approx 0.628 s.

Springs in series (end to end) carry the same force and their stretches add, so the pair is softer. Springs in parallel (side by side) stretch equally and their forces add. A block between two springs fixed to opposite walls is a parallel pair: moved by x, both push it back.

1keq=1k1+1k2\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2}Series. Two equal springs: k/2, so the period grows by √2.
keq=k1+k2k_{eq} = k_1 + k_2Parallel, or between two walls: T = 2π√(m/(k₁ + k₂)).

6The simple pendulum

A small bob on a light string of length ll, pulled aside by θ\theta: its weight mgmg has a part mgcos⁡θmg\cos\theta along the string and a part mgsin⁡θmg\sin\theta along the arc, back toward the bottom. That is the restoring force. For small angles sin⁡θ≈θ\sin\theta \approx \theta (in radians; at 10°, 0.1745 rad against 0.1736) and θ=x/l\theta = x/l, where xx is the distance along the arc:

F≈−mgθ=−mgl xF \approx -mg\theta = -\frac{mg}{l}\,xLike a spring with k = mg/l, so ω² = g/l: the mass cancels.
T=2πlgT = 2\pi\sqrt{\frac{l}{g}}
  • The period depends only on the length and g: not on the mass, and not on the amplitude (for small swings).
  • 4 times longer: twice the period.
  • For T=2T = 2 s with g=10g = 10 m/s²: l=gT2/4π2=40/39.5≈1.01l = gT^2/4\pi^2 = 40/39.5 \approx 1.01 m.
  • On the Moon (g/6g/6): T′=6 T≈2.45 TT' = \sqrt{6}\,T \approx 2.45\,T.

7More oscillators, and phase

T=2πImgdT = 2\pi\sqrt{\frac{I}{mgd}}Physical pendulum: a rigid body swinging about a pivot; I about the pivot, d from pivot to centre of mass.
T=2πICT = 2\pi\sqrt{\frac{I}{C}}Torsional pendulum: a disc twisting on a wire with twisting constant C.

A uniform rod of length LL swinging about one end: I=mL2/3I = mL^2/3, d=L/2d = L/2, so T=2π2L/3gT = 2\pi\sqrt{2L/3g}.

The phase difference between two oscillations with the same ω\omega is Δϕ=ϕ1−ϕ2\Delta\phi = \phi_1 - \phi_2.

Phase differenceWhat you see
00 or 2π2\piin phase: they move together
π\piout of phase: always opposite (one at the right end when the other is at the left end)
π/2\pi/2one passes the centre as the other reaches an end

Summary

Key ideas

  • SHM is periodic, oscillatory motion with a restoring force proportional to the displacement: F = −kx.
  • The shadow of a point moving round a circle does SHM, so x = A sin(ωt + φ).
  • A is the amplitude, ω the angular frequency, ωt + φ the phase and φ the phase at t = 0.
  • The speed is largest, Aω, at the centre; the acceleration is largest, Aω², at the ends.
  • a = −ω²x is the test for SHM.
  • T = 2π/ω, f = 1/T and ω = 2πf; never T = 2π/f.
  • A spring and mass: T = 2π√(m/k), independent of the amplitude.
  • Springs in series are softer; springs in parallel (or between two walls) add their constants.
  • A simple pendulum: T = 2π√(l/g) for small swings, independent of the mass.
  • In accelerating lifts and cars use the effective g.
  • Physical pendulum T = 2π√(I/mgd); torsional pendulum T = 2π√(I/C).
  • Phase difference 0 means in step, π means opposite, π/2 means a quarter cycle apart.

Every equation

Restoring force
F=−kxF = -kx
Displacement
x=Asin⁡(ωt+ϕ)x = A\sin(\omega t + \phi)
Velocity
v=Aωcos⁡(ωt+ϕ)v = A\omega\cos(\omega t + \phi)
Speed at x
v=ωA2−x2v = \omega\sqrt{A^2 - x^2}
Acceleration
a=−ω2xa = -\omega^2 x
Largest speed
vmax=Aωv_{max} = A\omega
Largest acceleration
amax=Aω2a_{max} = A\omega^2
Period
T=2π/ωT = 2\pi/\omega
Frequency
f=1/T=ω/2πf = 1/T = \omega/2\pi
Spring
T=2πm/kT = 2\pi\sqrt{m/k}
Spring frequency
f=12πk/mf = \tfrac{1}{2\pi}\sqrt{k/m}
Series
1/keq=1/k1+1/k21/k_{eq} = 1/k_1 + 1/k_2
Parallel
keq=k1+k2k_{eq} = k_1 + k_2
Pendulum
T=2πl/gT = 2\pi\sqrt{l/g}
Car accelerating
geff=g2+a2g_{eff} = \sqrt{g^2 + a^2}
Lift
geff=g±ag_{eff} = g \pm a
Physical pendulum
T=2πI/mgdT = 2\pi\sqrt{I/mgd}
Torsional pendulum
T=2πI/CT = 2\pi\sqrt{I/C}
Phase difference
Δϕ=ϕ1−ϕ2\Delta\phi = \phi_1 - \phi_2

Previous year questions with solutions

Real JEE and NEET questions on simple harmonic motion (shm). Try each one before you open the solution.

Q1NEET 2026One correct option

Savitha, a XI standard student, while conducting an experiment to determine the effective length of a simple pendulum LL, notes down the data of time taken to complete 30 oscillations as 60 s and hence calculates the length of the simple pendulum as: (Take π2=9.8{\pi}^{2}=9.8, and g=9.8m/s2g=9.8 m/s^{2} )

  1. A0.75 m
  2. B1.5 m
  3. C2 m
  4. D1 m
Show answer and solution

Answer: Option D

One oscillation takes T=6030=2 sT = \dfrac{60}{30} = 2\ \mathrm{s}. Then

L=gT24π2=9.8×224×9.8=1 mL = \dfrac{gT^{2}}{4\pi^{2}} = \dfrac{9.8 \times 2^{2}}{4 \times 9.8} = 1\ \mathrm{m}

A pendulum with a period of 2 s2\ \mathrm{s} — one second for each swing from side to side — is called a seconds pendulum, and it is about a metre long.

The trap is C, 2 m2\ \mathrm{m}, which squares the π\pi in 2π2\pi but not the 22, dividing by 2π22\pi^{2} instead of 4π24\pi^{2}. The others simply do not fit: with π2=g\pi^{2} = g, T=2LT = 2\sqrt{L}, so 0.75 m0.75\ \mathrm{m} gives 1.73 s1.73\ \mathrm{s} and 1.5 m1.5\ \mathrm{m} gives 2.45 s2.45\ \mathrm{s}, not 2 s2\ \mathrm{s}.

Q2JEE Main 2026One correct option

A simple pendulum of string length 30 cm performs 20 oscillations in 10 s . The length of the string required for the pendulum to perform 40 oscillations in the same time duration is

____\_\_\_\_ cm . [Assume that the mass of the pendulum remains same.]

  1. A0.75
  2. B7.5
  3. C15
  4. D120
Show answer and solution

Answer: Option B

First the two periods. Twenty oscillations in 10 s10\ \mathrm{s} give T1=0.5 sT_{1} = 0.5\ \mathrm{s}; forty in the same time give T2=0.25 sT_{2} = 0.25\ \mathrm{s}. The period must halve.

Since T∝LT \propto \sqrt{L}, L∝T2L \propto T^{2}:

L2=L1(T2T1)2=30×(12)2=7.5 cmL_{2} = L_{1}\left(\dfrac{T_{2}}{T_{1}}\right)^{2} = 30 \times \left(\dfrac{1}{2}\right)^{2} = 7.5\ \mathrm{cm}

The trap is C, 15 cm15\ \mathrm{cm}, which halves the length as if T∝LT \propto L. D, 120 cm120\ \mathrm{cm}, goes the wrong way: a longer pendulum swings more slowly. A, 0.75 cm0.75\ \mathrm{cm}, is the right idea with a slip of a factor of ten.

Q3JEE Advanced 2016One or more correct options

A block with mass M is connected by a massless spring with stiffness constant k to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude A about an equilibrium position x₀. Consider two cases:

(i) when the block is at x₀; and

(ii) when the block is at x = x₀ + A.

In both cases, a particle with mass m( < M) is softly placed on the block after which they stick on each other. Which of the following statement(s) is(are) true about the motion after the mass m is placed on the mass M?

  1. AThe amplitude of oscillation in the first case changes by a factor of Mm+M\sqrt{\frac{M}{m+M}}, whereas in the second case it remains unchanged.
  2. BThe final time period of oscillation in both the cases is same
  3. CThe total energy decreases in both the cases
  4. DThe instantaneous speed at x₀ of the combined masses decreases in both the cases
Show answer and solution

Answer: Options A, B, D

Case (i), at x0x_{0}: the block is moving at AωA\omega, so momentum is shared, MAω=(M+m)v′MA\omega = (M + m)v', and with ω′=kM+m\omega' = \sqrt{\dfrac{k}{M + m}} the amplitude becomes AMM+mA\sqrt{\dfrac{M}{M + m}}. Case (ii), at x0+Ax_{0} + A: the block is at rest at an extreme, so the amplitude stays AA. Statement A is true.

B is true: in both cases the mass is M+mM + m on the same spring, so T=2πM+mkT = 2\pi\sqrt{\dfrac{M + m}{k}}.

C is false. In case (i) the sticking is inelastic and energy is lost, but in case (ii) the energy 12kA2\dfrac{1}{2}kA^{2} is unchanged, since AA is.

D is true. In case (i) the speed at x0x_{0} is v′=MM+mAωv' = \dfrac{M}{M + m}A\omega, less than AωA\omega. In case (ii) it is Aω′A\omega', and ω′<ω\omega' < \omega.

The trap is C: energy is lost only when there is motion to share at the moment of sticking.

Practice questions, easy to hard

Three questions from the simple harmonic motion (shm) practice ladder: one easy, one medium, one hard.

Q4Numerical answer

The constant ω\omega in a=−ω2xa = -\omega^{2}x is the angular frequency of the SHM, in rad/s. It fixes how fast the motion repeats:

ω=2πT=2πf\omega = \dfrac{2\pi}{T} = 2\pi f

Here TT is the period in seconds, and f=1Tf = \dfrac{1}{T} is the frequency: the number of oscillations each second, in hertz (Hz). Keep the three apart: ω\omega in rad/s, ff in Hz, TT in s.

For example, a=−16xa = -16x means ω2=16\omega^{2} = 16, so ω=4 rad/s\omega = 4\ \mathrm{rad/s}, T=2π4=π2≈1.57 sT = \dfrac{2\pi}{4} = \dfrac{\pi}{2} \approx 1.57\ \mathrm{s} and f=1T≈0.64 Hzf = \dfrac{1}{T} \approx 0.64\ \mathrm{Hz}.

A particle moves along a line with a=−100xa = -100x (SI units). How many oscillations does it make each second? Give the frequency ff in Hz, to two decimal places.

Show answer and solution

Answer: 1.5915 Hz

ω2=100\omega^{2} = 100, so ω=10 rad/s\omega = 10\ \mathrm{rad/s}. Then f=ω2π=102π≈1.59 Hzf = \dfrac{\omega}{2\pi} = \dfrac{10}{2\pi} \approx 1.59\ \mathrm{Hz}.

The trap is answering 1010: that is ω\omega, in rad/s, not the number of oscillations per second. Another is 100100, which forgets the square: the 100100 is ω2\omega^{2}. And T=2π10≈0.63 sT = \dfrac{2\pi}{10} \approx 0.63\ \mathrm{s} is the period, the time for one oscillation, not the number per second.

Q5One or more correct options

The reverse can happen too: a piece falls OFF a block. If nothing pushes it, the piece leaves with the velocity it had, and so does the rest: no momentum passes between them. The remaining block carries on from the same place at the same speed, but with a new ω\omega, because its mass has changed.

A block oscillating on a horizontal spring with amplitude AA and period TT is at an extreme when three quarters of its mass drops off, leaving one quarter attached. Which statements are true?

  1. AThe amplitude stays AA
  2. BThe period becomes T2\dfrac{T}{2}
  3. CThe greatest speed doubles
  4. DThe energy of the oscillation falls to a quarter
Show answer and solution

Answer: Options A, B, C

At the extreme the block is at rest, so what remains starts from rest at the same stretch: the amplitude stays AA (A). A quarter of the mass halves the period, T∝massT \propto \sqrt{\text{mass}} (B), so ω\omega doubles and vmax=Aωv_{max} = A\omega doubles too (C).

D is false: the energy 12kA2\dfrac{1}{2}kA^{2} depends only on kk and AA, and neither has changed. The piece that fell off was at rest, so it took no energy with it.

Q6Numerical answer

The momentum p=mvp = mv follows the speed, so from v2=ω2(A2−x2)v^{2} = \omega^{2}(A^{2} - x^{2}) a graph of pp against xx is an ellipse. It crosses the xx-axis at ±A\pm A and the pp-axis at ±pmax=±mAω\pm p_{max} = \pm mA\omega. At x=0x = 0 all the energy is kinetic, so E=pmax22mE = \dfrac{p_{max}^{2}}{2m}, and ω=pmaxmA\omega = \dfrac{p_{max}}{mA}.

The momentum–displacement graph of a 2 kg2\ \mathrm{kg} particle in SHM is an ellipse that crosses the xx-axis at ±0.5 m\pm 0.5\ \mathrm{m} and the pp-axis at ±4 kg m/s\pm 4\ \mathrm{kg\,m/s}. What is the particle's total energy, in joules?

Show answer and solution

Answer: 4 J

E=pmax22m=164=4 JE = \dfrac{p_{max}^{2}}{2m} = \dfrac{16}{4} = 4\ \mathrm{J}. (Also ω=pmaxmA=42×0.5=4 rad/s\omega = \dfrac{p_{max}}{mA} = \dfrac{4}{2 \times 0.5} = 4\ \mathrm{rad/s}, and 12mω2A2=12×2×16×0.25=4 J\dfrac{1}{2}m\omega^{2}A^{2} = \dfrac{1}{2} \times 2 \times 16 \times 0.25 = 4\ \mathrm{J} agrees.)

The trap is treating 4 kg m/s4\ \mathrm{kg\,m/s} as a speed, 12×2×42=16 J\dfrac{1}{2} \times 2 \times 4^{2} = 16\ \mathrm{J}: it is a momentum, so the speed is only 2 m/s2\ \mathrm{m/s}. Another is using the 0.5 m0.5\ \mathrm{m} on the xx-axis as if it were part of the energy at x=0x = 0, where the potential energy is zero.