SHM and uniform circular motion

This is a specific conceptual link inside the CBSE Class 11 Physics chapter "Oscillations," sitting right after SHM's kinematic equations $x=A\cos(\omega t+\phi)$, $v=-A\omega\sin(\omega t+\phi)$ and $a=-\omega^2x$ have been introduced and explicitly connected to uniform circular motion, exactly as prescribed. The expectation at board level is to be able to state and use the projection idea, derive $x(t)$ from the geometry of the reference circle, and apply it to phase-difference questions — not to study circular motion itself in depth, nor to touch anything beyond this prescription. Free, forced or damped oscillations and resonance are explicitly out of the current 2025-26 syllabus and are not developed here; where the analogy would naturally lead there (for instance, "what if the circular speed itself changes?"), this article stops at a one-line pointer to higher classes rather than deriving it.

Set up a circle of radius $A$ centred at the origin O, and let a particle P move around it at constant angular speed $\omega$, so its angular position at time $t$ is $\theta(t)=\omega t+\phi$, with $\phi$ the angle at $t=0$. This is uniform circular motion in the ordinary sense: P's speed along the circle, $A\omega$, never changes, and neither does $\omega$ itself. Now drop a perpendicular from P to the x-axis and call the foot of that perpendicular M. As P sweeps around the circle once every period $T=2\pi/\omega$, M slides back and forth along the x-axis between $-A$ and $+A$, and its position is

where: $A$ is the radius of the reference circle (the SHM amplitude), $\omega$ is P's angular speed on the circle (the SHM angular frequency), $\phi$ is P's angular position at $t=0$ (the SHM phase constant), and $\omega t+\phi$ is P's instantaneous angle, called the phase of the motion at time $t$. Physically, M's position is simply the horizontal component of P's position vector — as P sweeps through angles where $\cos\theta$ is near $1$ (close to the positive x-axis), M sits near $+A$ and barely moves; as P crosses the top or bottom of the circle, where $\cos\theta$ changes fastest, M rushes through the centre at its highest speed. This is exactly the "slow near the extremes, fast at the centre" behaviour of SHM, now visible as ordinary circular geometry rather than a fact you have to memorise from a graph.

The same particle P also has a shadow on the vertical diameter, at height $y(t)=A\sin(\omega t+\phi)$ — an SHM of the same amplitude and angular frequency as $x(t)$, but exactly a quarter-cycle out of step with it, since $\sin\theta$ and $\cos\theta$ always differ by $90°$. This single picture is why the "reference circle" is such a powerful tool: two SHMs are in phase precisely when their reference points sit at the same angle on the circle at every instant, and out of phase by an angle $\theta$ when the two reference points are permanently separated by $\theta$ on the circle. What is a difficult algebraic comparison of two trigonometric expressions becomes an easy comparison of two angles on a clock face.

The reference-circle picture is not limited to displacement — it hands you the SHM velocity and acceleration formulas almost without calculation. P moves on the circle at constant speed $A\omega$, directed tangent to the circle (perpendicular to OP). The horizontal component of this tangential velocity — the rate at which the shadow M moves along the x-axis — is

where: the negative sign and the sine (rather than cosine) arise because the tangential velocity vector is perpendicular to OP, so its horizontal component is $-A\omega$ times the sine of the same angle that gave $x$ as $A\omega$ times the cosine. Likewise, P undergoes centripetal acceleration of constant magnitude $A\omega^2$, always directed from P back toward the centre O. Its horizontal component — the acceleration of the shadow M — is $-A\omega^2\cos(\omega t+\phi)$, which is exactly

where: the acceleration of the shadow is always directed toward O's projection, the mean position, and its size is proportional to how far the shadow currently is from that mean position — the defining signature of SHM, obtained here purely from the fact that centripetal acceleration always points toward the circle's centre.

The projection recovers the SHM restoring-force law too, though only as a consistency check here, since $F=-kx$ for the spring is established independently in the parent SHM topic. If a real spring-mass system of angular frequency $\omega=\sqrt{k/m}$ is imagined as the shadow of a particle on a reference circle of radius equal to the amplitude, then $a=-\omega^2x$ from the circle picture matches $a = F/m = -kx/m = -(k/m)x$ from Newton's second law term for term, since $\omega^2 = k/m$. The two descriptions — force-based and circle-based — are two routes to the identical equation of motion, which is exactly why the reference-circle method is trustworthy and not merely a mnemonic.

Board level - Displacement of a projected point at a given instant

A particle moves on a reference circle of radius $0.5$ m with a constant angular speed of $4$ rad/s, starting from the point on the circle at angle $30°$ to the x-axis at $t=0$. Write the equation of the SHM traced by its projection on the x-axis, and find the displacement at $t=1$ s.

Step 1. Identify the reference-circle quantities and convert to SI/radians. $A = 0.5$ m, $\omega = 4$ rad/s, and $\phi = 30° = \pi/6$ rad $\approx 0.524$ rad (the conversion to radians is essential because $\omega t$ is automatically in radians).

Step 2. Write the SHM equation using the dictionary $x(t)=A\cos(\omega t+\phi)$:

Step 3. Substitute $t=1$ s. The phase angle is $4(1)+\pi/6 = 4 + 0.524 = 4.524$ rad. Since $4.524$ rad lies between $\pi\approx 3.14$ rad and $3\pi/2\approx4.71$ rad, the reference point P is in the third quadrant of the circle, where cosine is negative, so the shadow should come out on the negative side of the mean position.

Step 4. Evaluate. $\cos(4.524\ \text{rad}) \approx -0.183$, so $x = 0.5\times(-0.183) \approx -0.0915$ m. Answer: $x(t)=0.5\cos(4t+\pi/6)$ m, and at $t=1$ s the shadow is about $9.2$ cm on the negative side of the mean position, matching the quadrant prediction from Step 3.

Competitive level - Phase relationship between two projections at 90°

Two particles execute SHM of the same amplitude and angular frequency, obtained as projections of two points moving on the same reference circle but separated by a constant angle of $90°$. If one particle's displacement is $x_1 = A\cos(\omega t)$, write the displacement of the second particle, describe the phase relationship, and find the ratio of their velocities at the instant $x_1$ is at its maximum.

Step 1. Locate the second reference point. Since the second point leads the first by a constant $90° = \pi/2$ rad on the circle at all times, its angular position is $\omega t + \pi/2$ wherever the first point's is $\omega t$; projecting onto the x-axis, $x_2 = A\cos(\omega t + \pi/2) = -A\sin(\omega t)$ (using the standard identity $\cos(\theta+\pi/2)=-\sin\theta$).

Step 2. Describe the phase relationship geometrically. The two dots sit permanently $90°$ apart on the same circle, so their shadows are two SHMs of identical $A$ and $\omega$ that are out of phase by a quarter cycle: whenever $x_1$ reaches an extreme (P at angle $0°$ or $180°$), the second point sits at $90°$ or $270°$, exactly where its shadow $x_2$ crosses zero — and vice versa.

Step 3. Differentiate to get both velocities (or project both particles' tangential speeds $A\omega$, as in the physical picture above): $v_1 = -A\omega\sin(\omega t)$ and $v_2 = -A\omega\cos(\omega t)$.

Step 4. Evaluate at the instant $x_1$ is maximum, i.e. $\omega t = 0$. Then $v_1 = -A\omega\sin(0) = 0$ (the particle momentarily stops at the extreme, consistent with the reference point P sitting exactly at the top of the projection axis, where its whole tangential velocity is vertical, contributing nothing to $x$), while $v_2 = -A\omega\cos(0) = -A\omega$, its maximum possible speed. Answer: $x_2 = -A\sin(\omega t)$, the two SHMs are $90°$ out of phase, and at the instant $x_1$ is maximum the ratio $v_1:v_2 = 0:(-A\omega) = 0$ — particle 1 is momentarily at rest while particle 2 passes through its mean position at top speed. Timing note: in an exam, do not re-derive $v_1$ and $v_2$ by differentiation if time is short — go straight to the reference-circle picture: at $\omega t=0$, particle 1's reference point is at the "top" of the projection axis (velocity purely perpendicular to the x-axis, so $v_1=0$), while particle 2's reference point (a further $90°$ ahead) is crossing the axis itself (velocity purely along the x-axis, so $|v_2|=A\omega$, the maximum). Reading the geometry directly like this typically saves 30-45 seconds per phase-difference question over writing and differentiating both equations from scratch.

• Thinking the particle undergoing SHM is itself moving in a circle. It only ever moves along the straight-line diameter; the circle is a geometric construction used to derive and visualize the motion, not a physical path the oscillating mass travels.
• Forgetting that the reference particle's angular speed on the circle and the SHM's angular frequency $\omega$ are numerically identical — this identity is exactly why $\omega$ carries the name "angular frequency" in SHM even though nothing physically rotates in a spring-mass system.
• Mixing up sine versus cosine when projecting onto different axes: the axis from which the phase angle $\phi$ is measured gives $A\cos(\omega t+\phi)$, while the perpendicular axis gives $A\sin(\omega t+\phi)$, a version shifted by a quarter cycle relative to the first.
• Forgetting to convert the starting angle $\phi$ from degrees to radians before adding it to $\omega t$ — since $\omega t$ is automatically dimensionless (radians), mixing degrees and radians inside one cosine gives a badly wrong numerical answer even when the method is correct.
• Assuming the shadow's speed is constant because the real particle's speed on the circle is constant — the projection's speed changes continuously, fastest at the mean position and zero at the extremes, precisely the opposite of "uniform."

• Describe how projecting a particle in uniform circular motion onto a diameter produces SHM, and identify which quantity on the circle corresponds to which quantity in the SHM equation.
• Derive $x(t)=A\cos(\omega t+\phi)$ starting from the geometry of the reference circle, without assuming the SHM equation in advance.
• Explain, using the reference circle, why $\omega$ is called angular frequency in SHM even though the real oscillator never rotates.
• Use the reference-circle picture to read off the velocity $v=-A\omega\sin(\omega t+\phi)$ and acceleration $a=-\omega^2x$ without differentiating from scratch.
• Determine the phase relationship between two SHMs of the same amplitude and angular frequency from the angular separation of their reference points on a shared circle.
• State clearly that the SHM particle's speed is not constant even though the corresponding reference particle's circular speed is.

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