Q.Starting from Hooke's law for the restoring force of a spring, $F = -kx$, derive the differential equation of motion for a particle executing S.H.M. Show that $x(t) = A\cos(\omega t + \phi)$ is a solution of this equation, and state how $\omega$ is related to $k$ and the mass $m$.

Showing the 12 most recent of 52 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.In SHM, the restoring force is directly proportional to

   (a) Velocity

   (b) Displacement

   (c) Acceleration

   (d) Time

   ›Reveal solutionSolution

   The defining equation of simple harmonic motion is F = -k x -- the restoring force is proportional to displacement x, with the negative sign showing it always acts to pull the object back toward equilibrium.

   Simple Harmonic Motion (SHM) is defined precisely by this relationship between restoring force and displacement:

   F = -k x

   where x is the displacement from the mean/equilibrium position and k is a positive constant (the force constant). This means:

   • The magnitude of the restoring force grows linearly with how far the object is displaced from equilibrium.
   • The negative sign shows the force always points back toward the equilibrium position, opposing the displacement.

   This is exactly what makes an oscillator's motion "simple harmonic" -- any motion where the restoring force is proportional to displacement (and directed opposite to it) will be SHM, with acceleration a = -(k/m)x also being proportional to displacement (which is why "acceleration" is sometimes tempting to pick, but the FUNDAMENTAL defining proportionality, stated in the force law itself, is with displacement).

   ✓Final answer

   (b) Displacement.

2. CBSE 2026Set ANNUAL1 markMCQ
   Q.In SHM, the maximum velocity occurs at

   (a) The equilibrium position

   (b) Half of the amplitude

   (c) Maximum displacement from equilibrium

   (d) It is constant throughout the motion

   ›Reveal solutionSolution

   Velocity in SHM is v(t) = -Aomegasin(omega t), which has its maximum magnitude (A*omega) exactly when displacement x = 0, i.e., at the equilibrium position.

   For SHM with displacement x(t) = A cos(omega t), the velocity is:

   v(t) = dx/dt = -A omega sin(omega t)

   The magnitude of velocity is maximum when |sin(omega t)| = 1, which occurs precisely when cos(omega t) = 0, i.e., when x = 0 -- the equilibrium (mean) position.

   At the extreme positions (x = +-A), sin(omega t) = 0, so velocity is zero there (the object momentarily stops before reversing direction) -- exactly the opposite of "maximum displacement." This makes physical sense energetically too: all the mechanical energy is kinetic at the mean position (zero elastic PE there) and all potential at the extremes (zero KE there), by conservation of energy in SHM.

   ✓Final answer

   (a) The equilibrium position.

3. CBSE 2026Set ANNUAL1 markMCQ
   Q.The restoring force in SHM is always

   (a) in the direction of motion

   (b) opposite to the direction of motion

   (c) independent of motion

   (d) zero

   ›Reveal solutionSolution

   The restoring force F = -kx always points opposite to the displacement x (toward equilibrium), which is why it is described as always acting opposite to the object's motion away from the mean position -- this restoring tendency is what continually pulls the oscillator back and sustains the oscillation.

   In SHM, F = -k x: whichever side of equilibrium the object is displaced to, the restoring force always points back toward the centre (equilibrium), i.e., in the direction opposite to the displacement vector. Practically, this means the restoring force is always directed so as to oppose (resist) the object's continued displacement away from the mean position -- described in this basic sense as "opposite to the direction of motion" away from equilibrium. (A more precise statement, taught alongside this idea, is that the restoring force is always opposite to displacement, not literally opposite to instantaneous velocity at every point of the cycle -- for example, on the way back toward the mean position, both the force and the velocity point in the same direction, toward equilibrium. But the basic, always-true relationship examined here is that the restoring force continually acts against the object's displacement/outward motion from equilibrium, which is why it restores the oscillation.)

   ✓Final answer

   (b) opposite to the direction of motion.

4. CBSE 2026Set ANNUAL1 markMCQ
   Q.In simple harmonic motion, what is the phase difference between displacement and velocity?

   (a) 3π/4

   (b) π

   (c) π/2

   (d) 0

   ›Reveal solutionSolution

   In simple harmonic motion, velocity is always π/2 (90°) out of phase with (ahead of) displacement.

   For SHM, x(t) = A sin(ωt + φ). Differentiating, v(t) = dx/dt = Aω cos(ωt + φ) = Aω sin(ωt + φ + π/2). Comparing the arguments of the sine functions for x and v, velocity has an extra phase of +π/2 relative to displacement. Physically: when displacement is maximum (at the extreme position), velocity is zero; when displacement is zero (at the mean position), velocity is maximum — this quarter-cycle offset is exactly a phase difference of π/2.

   ✓Final answer

   The correct option is (c) π/2.

5. CBSE 2026Set ANNUAL1 mark
   Q.Fill in the blank: Oscillatory motion is ________ only for small amplitude.

   ›Reveal solutionSolution

   General oscillatory motion behaves as simple harmonic motion (SHM) only when the amplitude is small.

   SHM requires a restoring force (or torque) that is exactly proportional to displacement from equilibrium, F = −kx. For many real oscillators — e.g. a simple pendulum, where the restoring torque involves sin θ — this proportionality is only approximate: sin θ ≈ θ holds accurately only for small angles θ (small amplitude). For larger amplitudes, the restoring force is no longer linear in the displacement; the motion is still periodic but is no longer strictly simple harmonic (the time period even starts depending slightly on the amplitude). This is why the familiar SHM formulas (like T = 2π√(l/g) for a pendulum) are valid only in the small-amplitude limit.

   ✓Final answer

   Oscillatory motion is simple harmonic only for small amplitude.

6. CBSE 2026Set ANNUAL1 mark
   Q.If the amplitude of a particle performing simple harmonic motion is doubled, then which quantity will become double?

   ›Reveal solutionSolution

   Doubling the amplitude of an SHM particle doubles its maximum velocity and maximum acceleration (both are ∝ A); the total energy, however, becomes four times as large, since E ∝ A².

   For a particle in SHM, x = A sin(ωt), v = Aω cos(ωt), a = −Aω² sin(ωt). The maximum values are x_max = A, v_max = Aω, and a_max = Aω². Since ω (angular frequency, set by the system's mass and restoring-force constant) does not depend on amplitude, both v_max and a_max are directly proportional to A — so if A is doubled, v_max and a_max each double as well. It's important to distinguish this from the total mechanical energy of the oscillator, E = ½mω²A², which is proportional to A² and therefore becomes four times as large (not double) when A is doubled.

   ✓Final answer

   The maximum velocity and maximum acceleration both become double; the total energy becomes four times (not double).

7. CBSE 2026Set ANNUAL1 markMCQ
   Q.The position of the body is given by x = A sin ωt. The time at which the displacement would be maximum is:

   (a) π/ω

   (b) π/2ω

   (c) πω

   (d) None of the above

   ›Reveal solutionSolution

   $x=A\sin\omega t$ reaches its peak value $A$ when the sine term equals 1, i.e. at $t=\pi/2\omega$.

   The position is $x(t)=A\sin\omega t$. Since $\sin\omega t$ ranges only between $-1$ and $+1$, the displacement $x$ is maximum (equal to $A$) exactly when $\sin\omega t = 1$.

   $\sin\theta = 1$ first occurs at $\theta = \pi/2$, so:

   $\omega t = \frac{\pi}{2} \quad\Rightarrow\quad t=\frac{\pi}{2\omega}$

   ✓Final answer

   (b) $\pi/2\omega$.

8. CBSE 2026Set ANNUAL1 mark
   Q.Write true or false: Every oscillatory motion is periodic but every periodic motion is not oscillatory.

   ›Reveal solutionSolution

   The statement is TRUE: 'oscillatory' is a stricter condition than 'periodic'.

   A periodic motion is any motion that repeats itself identically after a fixed time interval (the period) — for example, uniform circular motion (a particle going round and round a circle at constant speed) is periodic, but it is not oscillatory because the particle never reverses direction and moves back and forth about a mean position; it just keeps circulating the same way.

   An oscillatory motion, by contrast, is a to-and-fro motion about a fixed mean (equilibrium) position, repeated over and over — for example, a simple pendulum or a mass on a spring. Since this to-and-fro motion also repeats after a fixed interval, every oscillatory motion is automatically periodic. But as the circular-motion example shows, the converse is not true.

   ✓Final answer

   True — every oscillatory motion is periodic, but every periodic motion need not be oscillatory.

9. CBSE 2026Set ANNUAL1 markMCQ
   Q.A particle executes simple harmonic motion with a time period of 6 second and amplitude of 3 cm. Its maximum speed in cm/sec will be

   (a) 3π

   (b) 2π

   (c) π

   (d) π/2

   ›Reveal solutionSolution

   v_max = A omega = 3(2 pi/6) = pi cm/s. Answer (C).

   In SHM the maximum speed is v_max = A omega, where omega = 2 pi/T.

   Here A = 3 cm and T = 6 s, so:

   omega = 2 pi/6 = pi/3 rad/s.

   v_max = A omega = 3 x (pi/3) = pi cm/s.

   ✓Final answer

   (C) pi cm/s.

10. CBSE 2026Set ANNUAL1 markMCQ
    Q.The displacement of a particle moving in simple harmonic motion (SHM) at any instant is given by y = a cosωt. The acceleration after time t = T/4 (where T is time period) is

    (a) 0

    (b) − aω^2

    (c) aω^2

    (d) − aω

    ›Reveal solutionSolution

    At t = T/4 the displacement is zero (mean position), so acceleration = -omega^2 y = 0. Answer (A).

    Given y = a cos(omega t). Acceleration in SHM is:

    a(t) = d^2y/dt^2 = -a omega^2 cos(omega t) = -omega^2 y.

    At t = T/4: omega t = (2 pi/T)(T/4) = pi/2, and cos(pi/2) = 0.

    So y = 0 (the particle is at the mean position) and the acceleration = -omega^2 (0) = 0.

    ✓Final answer

    (A) 0.

11. CBSE 2026Set ANNUAL1 markMCQ
    Q.A simple harmonic oscillator has an amplitude a and time period T. The time required by it to travel from x = a to x = a/2 is

    (a) T/2

    (b) T/4

    (c) T/3

    (d) T/6

    ›Reveal solutionSolution

    Starting from the extreme (x = a), x = a/2 is reached at t = T/6. Answer (D).

    Start timing from the extreme position, so x = a cos(omega t) (at t = 0, x = a).

    When x = a/2:

    a/2 = a cos(omega t) -> cos(omega t) = 1/2 -> omega t = pi/3.

    Since omega = 2 pi/T:

    (2 pi/T) t = pi/3 -> t = (pi/3)(T/2 pi) = T/6.

    ✓Final answer

    (D) T/6.

12. CBSE 2025Set ANNUAL1 markMCQ
    Q.In simple harmonic motion, the phase difference between acceleration and displacement is (A) 0° (B) π (C) π/2 (D) 2π

    ›Reveal solutionSolution

    Acceleration and displacement in SHM are 180° (π radian) out of phase.

    For a particle in SHM, displacement is $x = A\sin(\omega t)$ and acceleration is:

    $a = -\omega^2x = -\omega^2A\sin(\omega t) = \omega^2A\sin(\omega t + \pi)$

    The negative sign means acceleration is always directed opposite to displacement (toward the mean position), which corresponds to a phase difference of $\pi$ radians (180°) between the two.

    ✓Final answer

    (B) π.