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SHM flash cards

Master SHM through 94 NEET-level recall cards, systematically structured one idea at a time. Revise concept-wise, identify the areas where you need improvement, and focus your preparation with greater precision.

SHM, question and answer

25 of this chapter's 94 cards, laid out open so you can read straight through. The remaining 69 are in the interactive deck, where the answer stays hidden until you commit to one.

  1. 1.What is periodic motion?

    Motion that repeats itself at regular intervals of time (e.g. motion of hands of a clock, planets around the Sun).

    Hint: Repeats after a fixed period.

  2. 2.What is oscillatory (vibratory) motion?

    To-and-fro motion of a body about a fixed mean (equilibrium) position. Every oscillatory motion is periodic, but not every periodic motion is oscillatory.

    Hint: To-and-fro about equilibrium.

  3. 3.Define Simple Harmonic Motion (SHM).

    Oscillatory motion in which the restoring force (or acceleration) is directly proportional to the displacement from the mean position and is always directed towards it: F=kxF = -kx.

    Hint: FxF \propto -x.

  4. 4.Write the defining differential equation of SHM.

    d2xdt2+ω2x=0\dfrac{d^2x}{dt^2} + \omega^2 x = 0, where ω2=k/m\omega^2 = k/m is the square of the angular frequency.

    Hint: Second-order, restoring term ω2x\omega^2 x.

  5. 5.What is the general displacement equation of a particle in SHM?

    x=Asin(ωt+ϕ)x = A\sin(\omega t + \phi) or equivalently x=Acos(ωt+ϕ)x = A\cos(\omega t + \phi), where AA is amplitude, ω\omega angular frequency and ϕ\phi initial phase.

    Hint: Sine/cosine of (ωt+ϕ)(\omega t + \phi).

  6. 6.What does the amplitude AA represent in SHM?

    The maximum displacement of the particle from its mean position. It fixes the energy of the oscillation but not the period.

    Hint: Peak displacement.

  7. 7.Define the time period TT of SHM and relate it to ω\omega.

    TT is the time for one complete oscillation. T=2πωT = \dfrac{2\pi}{\omega}.

    Hint: 2π/ω2\pi/\omega.

  8. 8.Define frequency ff and relate it to ω\omega and TT.

    ff is the number of oscillations per second. f=1T=ω2πf = \dfrac{1}{T} = \dfrac{\omega}{2\pi}, unit hertz (Hz).

    Hint: 1/T=ω/2π1/T = \omega/2\pi.

  9. 9.What is angular frequency ω\omega?

    ω=2πf=2πT\omega = 2\pi f = \dfrac{2\pi}{T}; it measures the rate of phase change, unit rad s1^{-1}.

    Hint: 2πf2\pi f.

  10. 10.What is the phase (ωt+ϕ)(\omega t + \phi) of SHM?

    The quantity that determines the state (position and direction of motion) of the oscillator at a given instant. ϕ\phi is the phase constant (initial phase / epoch).

    Hint: State of oscillator at time tt.

  11. 11.Write the velocity of a particle in SHM as a function of time (for x=Asinωtx=A\sin\omega t).

    v=dxdt=Aωcosωtv = \dfrac{dx}{dt} = A\omega\cos\omega t; maximum speed vmax=Aωv_{max} = A\omega at the mean position.

    Hint: Differentiate xx.

  12. 12.Express velocity in SHM in terms of displacement xx.

    v=ωA2x2v = \omega\sqrt{A^2 - x^2}. It is maximum (AωA\omega) at x=0x=0 and zero at x=±Ax=\pm A.

    Hint: ωA2x2\omega\sqrt{A^2-x^2}.

  13. 13.Write the acceleration of a particle in SHM (for x=Asinωtx=A\sin\omega t).

    a=d2xdt2=Aω2sinωt=ω2xa = \dfrac{d^2x}{dt^2} = -A\omega^2\sin\omega t = -\omega^2 x; maximum magnitude amax=Aω2a_{max} = A\omega^2 at the extremes.

    Hint: ω2x-\omega^2 x.

  14. 14.Where in SHM are velocity and acceleration each maximum/minimum?

    Velocity is maximum at the mean position and zero at extremes; acceleration is zero at the mean position and maximum at the extremes.

    Hint: They swap roles at mean vs extreme.

  15. 15.What is the phase difference between displacement and velocity in SHM?

    Velocity leads displacement by π2\dfrac{\pi}{2} (90°).

    Hint: π/2\pi/2 ahead.

  16. 16.What is the phase difference between displacement and acceleration in SHM?

    Acceleration is out of phase with displacement by π\pi (180°); a=ω2xa=-\omega^2 x.

    Hint: Opposite direction, π\pi.

  17. 17.Show the shape of the velocity–displacement curve in SHM.

    From v=ωA2x2v=\omega\sqrt{A^2-x^2}, we get v2A2ω2+x2A2=1\dfrac{v^2}{A^2\omega^2}+\dfrac{x^2}{A^2}=1: an ellipse.

    Hint: Ellipse.

  18. 18.Show the shape of the acceleration–displacement curve in SHM.

    Since a=ω2xa=-\omega^2 x, it is a straight line through the origin with negative slope ω2-\omega^2.

    Hint: Straight line, slope ω2-\omega^2.

  19. 19.How is SHM related to uniform circular motion?

    SHM is the projection of uniform circular motion on a diameter of the reference circle. The radius equals the amplitude and the angular speed equals ω\omega.

    Hint: Projection of the reference circle.

  20. 20.What is the restoring force in SHM?

    The force that always acts towards the mean position and is proportional to displacement: F=kxF=-kx; kk is the force constant.

    Hint: kx-kx, towards mean.

  21. 21.Relate ω\omega, kk and mm for a linear SHM oscillator.

    ω=km\omega = \sqrt{\dfrac{k}{m}}, so T=2πmkT = 2\pi\sqrt{\dfrac{m}{k}}.

    Hint: k/m\sqrt{k/m}.

  22. 22.Write the kinetic energy of a particle in SHM as a function of xx.

    KE=12mω2(A2x2)KE = \dfrac{1}{2}m\omega^2(A^2 - x^2); maximum at the mean position.

    Hint: 12mω2(A2x2)\tfrac12 m\omega^2(A^2-x^2).

  23. 23.Write the potential energy of a particle in SHM as a function of xx.

    PE=12mω2x2=12kx2PE = \dfrac{1}{2}m\omega^2 x^2 = \dfrac{1}{2}kx^2; maximum at the extremes.

    Hint: 12kx2\tfrac12 kx^2.

  24. 24.Write the total mechanical energy in SHM.

    E=KE+PE=12mω2A2=12kA2E = KE + PE = \dfrac{1}{2}m\omega^2 A^2 = \dfrac{1}{2}kA^2; it is constant and independent of xx.

    Hint: 12mω2A2\tfrac12 m\omega^2 A^2, constant.

  25. 25.How does the total energy of SHM depend on amplitude and frequency?

    EA2E \propto A^2 and Eω2E \propto \omega^2 (i.e. f2\propto f^2). Doubling the amplitude quadruples the energy.

    Hint: EA2ω2E\propto A^2\omega^2.

Open the interactive deck for the other 69 cards, with self-grading so the ones you keep missing come back.

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