Velocity and acceleration in SHM

Velocity and acceleration in SHM

In plain words: a particle in SHM is like a child on a swing. It whizzes through the middle and stops for an instant at each end. The restoring acceleration along the motion is strongest at the ends and zero in the middle. For a real swing this describes the tangential component; its radial centripetal acceleration need not vanish at the bottom.

Velocity from the reference circle

The reference particle P moves anticlockwise round a circle of radius A with angular speed ω, so its speed is (NCERT Eq. 13.8)

the same rule as v = ωr for a turning body. Its velocity is along the tangent at P, at right angles to OP, so it makes the angle ωt + φ with the +y-axis (NCERT Fig. 13.11, page 266), and its x-component is −ωA sin(ωt + φ). That is the velocity of the projection P′:

The minus sign means that while sin(ωt + φ) is positive, P′ moves towards −x. The same result comes straight from calculus: v(t) = dx/dt (NCERT Eq. 13.10).

Speed and position. Square v(t), use sin^² = 1 − cos^² and put A cos(ωt + φ) = x:

So (a step beyond the NCERT text)

Take the + sign when the particle moves towards +x and the − sign when it moves towards −x: it passes each interior position twice per cycle, once in each direction, at the same speed. This gives speed ωA at x = 0 and zero at x = ±A, without needing the time.

Graphs against x (beyond the NCERT text). Since a = −ω^²x, the a–x graph is a straight line through the origin with slope −ω^². Rearranging v^² = ω^²(A^² − x^²) gives x^²/A^² + v^²/(ωA)^² = 1, so the v–x graph is an ellipse with semi-axes A and ωA.

Acceleration

The reference particle has a centripetal acceleration of size (ωA)^²/A = ω^²A, pointing along PO towards the centre. Its x-component is the acceleration of P′ (NCERT Fig. 13.12, page 266; that figure reuses the labels of NCERT Fig. 13.11, so its arrow on the x-axis is marked v(t), where it stands for the acceleration a(t) of P′):

This is NCERT Eq. 13.11; it also follows from a(t) = dv/dt (NCERT Eq. 13.12). Multiplied by the mass, it gives the force law for SHM.

So in SHM the acceleration is proportional to the displacement and opposite to it. When x > 0, a < 0; when x < 0, a > 0. For nonzero displacement, acceleration points towards the mean position; at x = 0 it vanishes (pages 266–267).

Note that the acceleration is not constant: it changes with x. The equations of motion for constant acceleration do not apply to SHM.

Displacement, velocity and acceleration together

With φ = 0 (page 267):

• x(t) = A cos ωt
• v(t) = −ωA sin ωt
• a(t) = −ω^²A cos ωt

All quantities vary sinusoidally with time; only their maxima differ and the different plots differ in phase.

NCERT Physics Part II (Class 11), Section 13.5, page 267

NCERT gives the phase differences as π/2 and π (page 267). “Ahead” follows from −sin θ = cos(θ + π/2) and −cos θ = cos(θ + π).

Solved examples

Common mistakes to avoid

Quick revision summary

• Displacement: x(t) = A cos(ωt + φ).
• Velocity: v(t) = −ωA sin(ωt + φ); speed ω√(A^² − x^²); greatest ωA at x = 0, zero at x = ±A.
• Acceleration: a(t) = −ω^²A cos(ωt + φ) = −ω^²x; zero at x = 0, greatest ω^²A at x = ±A; always towards the mean position.
• All three are sinusoidal with the same period; v is π/2 ahead of x, a is π/2 ahead of v, so a is π ahead of x (opposite to it).
• From the reference circle: the speed of P is ωA and its centripetal acceleration is ω^²A; their x-components give v and a of SHM.

Questions students ask

What are the average velocity and average speed in SHM over one period?

Over one complete period the average velocity of a particle in SHM is zero, because it returns to its starting point and its displacement is zero. Its average speed is the total distance, 4A, divided by the period: 4A/T = 2ωA/π, about 0.64 times the greatest speed ωA. For an amplitude of 3 cm and a period of 0.6 s, the average speed is 12 cm/0.6 s = 20 cm/s, while the greatest speed is 2π × 3/0.6 ≈ 31.4 cm/s.

What is the shape of the velocity–acceleration graph in SHM?

The velocity–acceleration graph of a particle in SHM is an ellipse. Since v = −ωA sin(ωt + φ) and a = −ω^²A cos(ωt + φ), squaring and adding gives v^²/(ωA)^² + a^²/(ω^²A)^² = 1, an ellipse with semi-axes ωA along the velocity axis and ω^²A along the acceleration axis. Velocity and acceleration are a quarter of a period out of step, so each is zero when the other is largest; compare the velocity–displacement graph, also an ellipse, and the acceleration–displacement graph, a straight line of slope −ω^².

What are the formulas for maximum velocity and maximum acceleration in SHM in terms of frequency?

In terms of the frequency ν, the greatest speed in SHM is v_max = ωA = 2πνA and the greatest acceleration is a_max = ω^²A = 4π^²ν^²A, since ω = 2πν. For a given amplitude the acceleration grows as the square of the frequency, so fast vibrations have very large accelerations even at tiny amplitudes: a prong of a 256 Hz tuning fork vibrating with an amplitude of 0.5 mm has v_max ≈ 0.80 m/s but a_max ≈ 1290 m/s^², about 130 times g.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. A particle performs SHM with amplitude 5 cm and angular frequency 10 rad/s. Find its greatest speed and greatest acceleration.

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   v_max = ωA = 10 × 0.05 = 0.5 m/s, and a_max = ω^²A = 100 × 0.05 = 5 m/s^².

2. Q2. In an SHM with ω = 5 rad/s, find the acceleration when the particle is at x = +2 cm.

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   a = −ω^²x = −25 × 0.02 = −0.5 m/s^²: 0.5 m/s^² towards the mean position.

3. Q3. At which positions in SHM are (a) the speed and (b) the acceleration greatest and least?

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   (a) The speed is greatest at the mean position and zero at the extremes. (b) The acceleration is zero at the mean position and greatest in size at the extremes.