SHM and uniform circular motion

How is SHM related to uniform circular motion?

This shows that if P moves uniformly on a circle, its projection P′ on a diameter of the circle executes SHM.

NCERT Physics Part II (Class 11), Section 13.4, page 265

In plain words: watch a point going steadily round a circle, but look at it edge-on, so you only see how far left or right it is. What you see is a point sliding back and forth along a line, fast in the middle and slowing to a stop at each end. That edge-on view is exactly SHM.

Seeing it with a ball and a shadow

NCERT suggests a simple experiment (NCERT Fig. 13.9, page 264). Tie a ball to a string and whirl it in a horizontal circle at a steady angular speed. Now look at it from the side, at the level of the circle, or watch its shadow, cast by a distant lamp level with the circle, on a wall at right angles to the plane of the circle. The ball, or its shadow, seems to move to and fro along a straight line, with the centre of the circle as the midpoint. With parallel light rays, the shadow is the perpendicular projection onto a diameter normal to the viewing direction and is SHM. The ball itself still follows the circle; a nearby point lamp introduces perspective distortion.

The reference circle

Put the centre O of the circle at the origin (NCERT Fig. 13.10, page 265).

1. At t = 0 the radius OP makes an angle φ with the +x-axis.
2. P turns anticlockwise at angular speed ω, so in time t it turns through a further angle ωt. OP then makes the angle ωt + φ with the +x-axis.
3. The foot of the perpendicular from P to the x-axis is P′, and OP′ is the projection of OP. The position of P′ is x(t) = A cos(ωt + φ), the defining equation of SHM.

Any diameter works. Projecting on the y-axis instead gives (page 265)

This is also SHM, with the same amplitude and period, but its phase differs by π/2: since sin θ = cos(θ − π/2), the y-motion is a quarter of a period behind the x-motion.

Matching the symbols. The reference circle gives each constant of SHM a picture:

The same picture also gives the velocity and acceleration of SHM, which the next section of the chapter works out.

Clockwise motion and angles in radians

If P goes round clockwise, the angle of OP from the +x-axis decreases: it is φ − ωt. Since cos(−θ) = cos θ, the projection is x(t) = A cos(ωt − φ). So for clockwise motion the phase constant in NCERT Eq. 13.4 is −φ, not φ (see Examples 2 and 4 below). The y-projection, A sin(φ − ωt), is then a quarter of a period ahead of the x-motion, not behind.

Radians. NCERT adds a footnote (page 265): an angle is a ratio of arc to radius, so it has no dimensions, and an angle written as a plain number is understood to be in radians. A degree must always be shown, so sin(15°) is the sine of 15 degrees but sin(15) is the sine of 15 radians. In ωt + φ, with ω in rad/s, give φ in radians too.

The forces are not the same

In spite of this connection between circular motion and SHM, the force acting on a particle in linear simple harmonic motion is very different from the centripetal force needed to keep a particle in uniform circular motion.

NCERT Physics Part II (Class 11), Section 13.4, page 265

The centripetal force on P has a constant size, mω^²A, and always points to the centre, so its direction keeps turning. The force on a particle in SHM always lies along one line, and its size changes from zero at the mean position to its largest at the extremes. The two are still linked: the x-component of the centripetal force, −mω^²A cos(ωt + φ) = −mω^²x, has exactly the form of the SHM force on a particle of the same mass (NCERT uses this projection in the next section, Section 13.5). Nothing in a block on a spring corresponds to the other component.

Solved examples

Common mistakes to avoid

Quick revision summary

• The projection of uniform circular motion on any diameter is SHM.
• Reference circle: radius = amplitude A; angular speed of P = angular frequency ω; angle of OP at t = 0 = phase constant φ for the anticlockwise x-projection.
• Anticlockwise from angle φ: x(t) = A cos(ωt + φ) and y(t) = A sin(ωt + φ); the y-motion is π/2 behind.
• Clockwise from angle φ: x(t) = A cos(ωt − φ).
• The connection is geometric: the centripetal force on P is not the SHM force, though its x-component has the same form, −mω^²x.
• Angles written without a unit are in radians.

Questions students ask

Is the projection of non-uniform circular motion simple harmonic?

In general it is not; a full-circle reference motion produces the standard full-amplitude sinusoidal projection when its angular speed is constant. If the particle’s angular speed changes, the angle of its radius is no longer ωt + φ, so the foot of the perpendicular from it to the diameter, x = A cos θ(t), still swings between −A and +A but is not a sinusoidal function of time. A stone whirled in a vertical circle on a string is an example: it moves faster at the bottom than at the top, so its shadow on the ground, with the Sun overhead, oscillates (periodically, if every turn takes the same time) but not in SHM.

Is uniform circular motion a combination of two SHMs?

Yes, uniform circular motion is the combination of two simple harmonic motions along perpendicular diameters, with the same amplitude and frequency and a phase difference of π/2: x = A cos ωt and y = A sin ωt give x^² + y^² = A^², a circle traced at the steady angular speed ω. With unequal amplitudes the same two SHMs trace an ellipse, and with a phase difference of 0 or π they give a straight line, which is itself an SHM along a slanting line. Combining perpendicular SHMs is a step beyond the NCERT text; read the other way, it is the reference-circle result that each projection of uniform circular motion is an SHM.

How is the reference circle used to find the time between two positions in SHM?

The reference circle turns a time question in SHM into an angle question: above or below each of the two positions on the diameter, take the point of the circle that matches the direction of motion, and divide the angle the radius turns through between those points by ω, so t = (angle turned)/ω = (angle/360°) × T. For example, the particle is farther than A/2 from the centre on the +x side while the radius lies within 60° of the +x-axis, a turn of 120°, so it spends T/3 of each period there: 0.8 s for a period of 2.4 s. Adding the other side, it spends two-thirds of its time farther than A/2 from the mean position, because it moves slowly near the extremes.

Practice questions

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

1. Q1. A particle moves on a circle of radius 10 cm at 2 revolutions per second. What are the amplitude and angular frequency of its shadow’s motion on a diameter?

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   The amplitude is the radius, 10 cm, and ω = 2π × 2 = 4π rad/s, about 12.6 rad/s.

2. Q2. A reference particle moves anticlockwise. How are the motions of its projections on the x-axis and the y-axis related?

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   Both are SHMs with the same amplitude and period. y = A sin(ωt + φ) = A cos(ωt + φ − π/2), so the y-motion is π/2 (a quarter of a period) behind the x-motion. (For clockwise motion the y-motion is π/2 ahead.)

3. Q3. Is the force on the reference particle the same as the force on the particle in SHM? Explain.

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   No. The centripetal force on P has a constant size, mω^²A, and keeps changing direction. The SHM force acts along one line and changes size, being −mω^²x. Only the x-component of the centripetal force has this form.