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  3. Energy in Simple Harmonic Motion

Oscillations and Waves · JEE & NEET Physics

Energy in Simple Harmonic Motion: notes and previous year questions

KE = ½k(A² − x²), PE = ½kx², the constant total ½kA², energy in time, energy methods, springs and pendulums.

Energy in Simple Harmonic Motion in short

  • In SHM energy moves between kinetic and potential; with no friction the total stays constant.
  • KE = ½k(A² − x²): largest at the centre, zero at the ends.
  • PE = ½kx², the area under the F–x line: zero at the centre, largest at the ends.
  • The total E = ½kA² = ½mω²A² depends only on the amplitude and the spring; E ∝ A².

1Energy swaps

In SHM energy keeps changing form. At the ends the block stops, so it has no kinetic energy: all the energy is stored in the stretched or squashed spring as potential energy. At the centre the spring is relaxed and all the energy is kinetic. With no friction (no damping) the total never changes.

E=KE+PE=constantE = KE + PE = \text{constant}

Example: a block passing the centre with 3 J of kinetic energy has 3 J of potential energy at the end of its swing.

2Kinetic energy

KE=12mv2KE = \tfrac{1}{2}mv^2 and in SHM v2=ω2(A2−x2)v^2 = \omega^2(A^2 - x^2). Since k=mω2k = m\omega^2:

KE=12mω2(A2−x2)=12k(A2−x2)KE = \tfrac{1}{2}m\omega^2(A^2 - x^2) = \tfrac{1}{2}k(A^2 - x^2)
  • At the centre (x=0x = 0): KEmax=12kA2KE_{max} = \tfrac{1}{2}kA^2. At the ends (x=±Ax = \pm A): KE=0KE = 0.
  • m=1m = 1 kg, k=100k = 100 N/m, pulled 10 cm: E=12×100×0.01=0.5E = \tfrac{1}{2} \times 100 \times 0.01 = 0.5 J, so at the centre 12mv2=0.5\tfrac{1}{2}mv^2 = 0.5 and vmax=1v_{max} = 1 m/s (also Aω=0.1×10A\omega = 0.1 \times 10).
  • 8 J at the centre means 8×(1−14)=68 \times (1 - \tfrac{1}{4}) = 6 J at x=A/2x = A/2, not 4 J: the xx is squared.

3Potential energy

Stretching a spring slowly takes a force kxkx that grows in a straight line. The work done is the area under that line: a triangle of base xx and height kxkx, so 12×x×kx\tfrac{1}{2} \times x \times kx.

PE=12kx2=12mω2x2PE = \tfrac{1}{2}kx^2 = \tfrac{1}{2}m\omega^2 x^2
  • Zero at the centre, largest (12kA2\tfrac{1}{2}kA^2) at the ends.
  • Twice the stretch stores 4 times the energy: 0.1 J at 2 cm becomes 0.9 J at 6 cm.
  • To store 2 J in a spring with k=400k = 400 N/m: x2=4/400x^2 = 4/400, x=0.1x = 0.1 m.

4Total energy

Adding them, the x2x^2 terms cancel: 12k(A2−x2)+12kx2=12kA2\tfrac{1}{2}k(A^2 - x^2) + \tfrac{1}{2}kx^2 = \tfrac{1}{2}kA^2. The total does not depend on xx or on time, only on the amplitude and the spring.

E=12kA2=12mω2A2E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}m\omega^2 A^2

Against x, PE is an upward parabola, KE the same parabola upside down, and E a flat line. Double the amplitude and the energy is 4 times as big.

PositionxKEPE
Centre0E (100%)0
Halfway outA/23E/4 (75%)E/4 (25%)
Equal shareA/2≈0.71AA/\sqrt{2} \approx 0.71AE/2 (50%)E/2 (50%)
End±A\pm A0E (100%)
KE=PE at x=±A2KE = PE \text{ at } x = \pm\frac{A}{\sqrt{2}}From ½kx² = ½ × ½kA².

5Energy and time

With x=Asin⁡ωtx = A\sin\omega t, and using cos⁡2θ=12(1+cos⁡2θ)\cos^2\theta = \tfrac{1}{2}(1 + \cos 2\theta):

KE=Ecos⁡2ωt=E2(1+cos⁡2ωt)KE = E\cos^2\omega t = \tfrac{E}{2}(1 + \cos 2\omega t)
PE=Esin⁡2ωt=E2(1−cos⁡2ωt)PE = E\sin^2\omega t = \tfrac{E}{2}(1 - \cos 2\omega t)
  • Both swing at twice the frequency, 2f2f (the energy is the same on both sides of the centre), and always opposite. An oscillator at 5 Hz has its KE rising and falling at 10 Hz.
  • Over a cycle, ⟨KE⟩=⟨PE⟩=E/2=14mω2A2=π2mf2A2\langle KE\rangle = \langle PE\rangle = E/2 = \tfrac{1}{4}m\omega^2A^2 = \pi^2 m f^2 A^2.
P=Fv=−12kA2ωsin⁡2ωtP = Fv = -\tfrac{1}{2}kA^2\omega\sin 2\omega tPower: the spring gives energy to the block, then takes it back. Average over a cycle: zero.

6Using energy

Use energy for the speed at a given position, where KE = PE, or the amplitude: it needs no time or phase. Use x = A sin(ωt + φ) for anything at a given time, the phase, or the acceleration.

7Vertical springs and pendulums

A block hung on a spring stretches it by x0=mg/kx_0 = mg/k to a new equilibrium. It oscillates about that point with the same period 2πm/k2\pi\sqrt{m/k}. Measured from the new equilibrium, the spring's and gravity's energies together change exactly like 12ky2\tfrac{1}{2}ky^2, so the total is again 12kA2\tfrac{1}{2}kA^2. Do not add mghmgh a second time.

For a pendulum the potential energy is mghmgh with h=l(1−cos⁡θ)≈lθ2/2h = l(1 - \cos\theta) \approx l\theta^2/2. With x=lθx = l\theta this is 12(mg/l)x2\tfrac{1}{2}(mg/l)x^2: a spring with k=mg/lk = mg/l.

E=mgl(1−cos⁡θ0)≈12mglθ02E = mgl(1 - \cos\theta_0) \approx \tfrac{1}{2}mgl\theta_0^2
vbottom=2gl(1−cos⁡θ0)v_{bottom} = \sqrt{2gl(1 - \cos\theta_0)}True at any angle, even where the motion is not SHM. Small angles: v = θ₀√(gl) = Aω.

Summary

Key ideas

  • In SHM energy moves between kinetic and potential; with no friction the total stays constant.
  • KE = ½k(A² − x²): largest at the centre, zero at the ends.
  • PE = ½kx², the area under the F–x line: zero at the centre, largest at the ends.
  • The total E = ½kA² = ½mω²A² depends only on the amplitude and the spring; E ∝ A².
  • KE = PE at x = A/√2; at A/2 the split is 3/4 kinetic, 1/4 potential.
  • KE and PE each swing at twice the frequency, always opposite, and each averages E/2.
  • The spring's power averages zero over a cycle.
  • Energy gives speeds at a position and amplitudes quickly; the equation of motion gives anything at a time.
  • A perfectly sticky collision keeps momentum but loses energy: the putty halves E, so A falls by √2.
  • A vertical spring oscillates about its new equilibrium with E = ½kA².
  • A pendulum's energy is mgl(1 − cos θ₀) ≈ ½mglθ₀², like a spring with k = mg/l.

Every equation

Kinetic energy
KE=12k(A2−x2)KE = \tfrac{1}{2}k(A^2 - x^2)
Kinetic energy (ω)
KE=12mω2(A2−x2)KE = \tfrac{1}{2}m\omega^2(A^2 - x^2)
Potential energy
PE=12kx2=12mω2x2PE = \tfrac{1}{2}kx^2 = \tfrac{1}{2}m\omega^2x^2
Total energy
E=12kA2=12mω2A2E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}m\omega^2A^2
Spring constant
k=mω2k = m\omega^2
Equal share
x=A/2x = A/\sqrt{2}
Share at x
PE/E=(x/A)2PE/E = (x/A)^2
KE in time
KE=E2(1+cos⁡2ωt)KE = \tfrac{E}{2}(1 + \cos 2\omega t)
PE in time
PE=E2(1−cos⁡2ωt)PE = \tfrac{E}{2}(1 - \cos 2\omega t)
Averages
⟨KE⟩=⟨PE⟩=E/2\langle KE\rangle = \langle PE\rangle = E/2
Average KE with f
⟨KE⟩=π2mf2A2\langle KE\rangle = \pi^2 m f^2 A^2
Power
P=−12kA2ωsin⁡2ωtP = -\tfrac{1}{2}kA^2\omega\sin 2\omega t
Vertical spring stretch
x0=mg/kx_0 = mg/k
Pendulum height
h=l(1−cos⁡θ)≈lθ2/2h = l(1 - \cos\theta) \approx l\theta^2/2
Pendulum energy
E=mgl(1−cos⁡θ0)E = mgl(1 - \cos\theta_0)
Speed at the bottom
v=2gl(1−cos⁡θ0)v = \sqrt{2gl(1 - \cos\theta_0)}
Putty
A=A0/2A = A_0/\sqrt{2}

Previous year questions with solutions

Real JEE and NEET questions on energy in simple harmonic motion. Try each one before you open the solution.

Q1JEE Main 2026One correct option

The equation of motion of a particle is given by x=asin⁡(50t+π/3)x=a\sin (50t+\pi /3) cm. The particle will come to rest at time t1t_{1} and it will have zero acceleration at time t2t_{2}. The t1t_{1} and t2t_{2} respectively are ________.

  1. Aπ300 s,π75 s\frac{\pi }{300}\text{ s}, \frac{\pi }{75}\text{ s}
  2. Bπ75 s,π300 s\frac{\pi }{75}\text{ s}, \frac{\pi }{300}\text{ s}
  3. Cπ300 s,π25 s\frac{\pi }{300}\text{ s}, \frac{\pi }{25}\text{ s}
  4. Dπ50 s,π100 s\frac{\pi }{50}\text{ s}, \frac{\pi }{100}\text{ s}
Show answer and solution

Answer: Option A

Here ω=50 rad/s\omega = 50\ \mathrm{rad/s}, and the phase 50t+π350t + \dfrac{\pi}{3} starts at π3\dfrac{\pi}{3} and grows.

The velocity v=50acos⁡(50t+π3)v = 50a\cos\left(50t + \dfrac{\pi}{3}\right) is first zero when the phase reaches π2\dfrac{\pi}{2} (the particle at its extreme): 50t1=π2−π3=π650t_{1} = \dfrac{\pi}{2} - \dfrac{\pi}{3} = \dfrac{\pi}{6}, so t1=π300 st_{1} = \dfrac{\pi}{300}\ \mathrm{s}.

The acceleration, −2500asin⁡(50t+π3)-2500a\sin\left(50t + \dfrac{\pi}{3}\right) (here aa is the amplitude), is next zero when the phase reaches π\pi (the particle back at the mean position): 50t2=π−π3=2π350t_{2} = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}, so t2=π75 st_{2} = \dfrac{\pi}{75}\ \mathrm{s}. That is A.

The trap is B, the same two times swapped: rest comes at the extreme and zero acceleration at the mean position, and starting at phase π3\dfrac{\pi}{3} the particle reaches the extreme first. D ignores the initial phase π3\dfrac{\pi}{3} and swaps them as well. C's π25 s\dfrac{\pi}{25}\ \mathrm{s} is a whole period, after which the particle is back where it started, not at the mean position.

Q2NEET 2026One correct option

A cylindrical cork of uniform density floats in a liquid of density ρ1{\rho}_{1}. If the cork is depressed slightly and released, it oscillates harmonically with time period TT. If the same cork floats in another liquid of density ρ2{\rho}_{2}, then the similar oscillation has time period 2T2T. The value of ρ2/ρ1{\rho}_{2}/{\rho}_{1} is:

  1. A1/4
  2. B4
  3. C2
  4. D1/21/2
Show answer and solution

Answer: Option A

For a floating body, T=2πmρAgT = 2\pi\sqrt{\dfrac{m}{\rho Ag}}. The cork has the same mass and the same cross-section in both liquids, so T∝1ρT \propto \dfrac{1}{\sqrt{\rho}}:

T2T1=ρ1ρ2=2,ρ2ρ1=14\dfrac{T_{2}}{T_{1}} = \sqrt{\dfrac{\rho_{1}}{\rho_{2}}} = 2, \qquad \dfrac{\rho_{2}}{\rho_{1}} = \dfrac{1}{4}

which is A. In the lighter liquid the cork floats four times as deep, and T=2πhgT = 2\pi\sqrt{\dfrac{h}{g}} doubles.

The trap is B, 44, the ratio upside down: a denser liquid pushes back harder and gives a SHORTER period, so a longer period means a less dense liquid. D, 12\dfrac{1}{2}, forgets that TT goes as the square root. C, 22, makes both slips.

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 energy in simple harmonic motion 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.