Simple harmonic motion — phase space, energy diagrams, and examples

Question

For a particle in SHM, describe how displacement, velocity, acceleration, and energy vary with time and position. What are the key relationships between these quantities?

(JEE Main 2024 tested energy variation; NEET asks velocity at mean/extreme positions)

Solution — Step by Step

Displacement: $x = A\sin(\omega t + \phi)$

Velocity: $v = A\omega\cos(\omega t + \phi) = \omega\sqrt{A^2 - x^2}$

Acceleration: $a = -A\omega^2\sin(\omega t + \phi) = -\omega^2 x$

The defining property: acceleration is proportional to displacement and directed towards the mean position.

At mean position ( $x = 0$ ): velocity is maximum ( $v = A\omega$ ), acceleration is zero, KE is maximum, PE is minimum.

At extreme position ( $x = \pm A$ ): velocity is zero, acceleration is maximum ( $a = A\omega^2$ ), KE is zero, PE is maximum.

Kinetic energy: $KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

Potential energy: $PE = \frac{1}{2}m\omega^2 x^2$

Total energy: $E = \frac{1}{2}m\omega^2 A^2 = \text{constant}$

Total energy is constant throughout the motion — it depends only on amplitude and frequency, not on instantaneous position.

Spring-mass: $T = 2\pi\sqrt{\frac{m}{k}}$

Simple pendulum: $T = 2\pi\sqrt{\frac{l}{g}}$

Both are independent of amplitude (for small oscillations in the pendulum case).

flowchart TD
    A["SHM: x = A sin(ωt)"] --> B["At x = 0 (mean position)"]
    A --> C["At x = ±A (extreme position)"]
    B --> D["v = max = Aω<br/>a = 0<br/>KE = max, PE = min"]
    C --> E["v = 0<br/>a = max = Aω²<br/>KE = 0, PE = max"]
    A --> F["Total Energy = ½mω²A²<br/>CONSTANT throughout"]
    style F fill:#90EE90,stroke:#333

Why This Works

SHM is the simplest possible periodic motion. It arises whenever a system is displaced from a stable equilibrium and the restoring force is proportional to displacement ( $F = -kx$ ). This linear restoring force produces sinusoidal oscillations.

The energy exchange between KE and PE drives the oscillation — at the mean position, all energy is kinetic (maximum speed), and at the extremes, all energy is potential (momentarily at rest before reversal). The total remains constant because no energy enters or leaves the system.

Alternative Method

For finding velocity at any position without using the time equation: use $v = \omega\sqrt{A^2 - x^2}$ . This position-based formula is faster than finding the time first and then computing velocity. At $x = A/2$ : $v = \omega\sqrt{A^2 - A^2/4} = \frac{\sqrt{3}}{2}A\omega$ . This specific value ( $\frac{\sqrt{3}}{2}A\omega$ at half-amplitude) appears frequently in JEE.

Common Mistake

Students assume the time to go from mean to extreme position is $T/4$ (correct), and then assume the time to go from mean to half-amplitude ( $x = A/2$ ) is $T/8$ (wrong). The motion is sinusoidal, not uniform. The particle moves faster near the centre and slower near the extremes. The actual time to reach $A/2$ from the mean is $T/12$ , not $T/8$ . Use $x = A\sin(\omega t)$ and solve for $t$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next