Login Free Test Start Mock

Physics · Oscillations

SHM and Uniform Circular Motion

Simple harmonic motion has a hidden geometry. NCERT §13.4 shows that the projection of uniform circular motion onto a diameter of the circle is simple harmonic motion. A reference particle moving uniformly on a circle of radius $A$ with angular speed $\omega$ casts a shadow on a diameter, and that shadow obeys $x(t) = A\cos(\omega t + \varphi)$ exactly. This "circle of reference" turns the abstract symbols of SHM into pictures — the radius becomes the amplitude, the swept angle becomes the phase, the angular speed becomes the angular frequency, and the tangential and centripetal vectors project into the SHM's velocity and acceleration.

The shadow experiment

Tie a ball to the end of a string and whirl it in a horizontal plane about a fixed point at a constant angular speed. The ball performs uniform circular motion. Now lower your eye to the level of the plane and watch the ball edge-on. The circular path collapses to a line, and the ball appears to move to and fro along that line, with the centre of rotation as the midpoint. You could equally watch the ball's shadow on a wall set perpendicular to the plane of the circle. NCERT describes precisely this demonstration in §13.4.

What you are observing in either case is the motion of the ball on a diameter of the circle — the diameter that is normal to your line of sight. The claim of this section is sharp and exact: that to-and-fro motion of the shadow is not merely like simple harmonic motion; it is simple harmonic motion. The geometry of the circle supplies, for free, the cosine law, the amplitude, the phase and the angular frequency that simple harmonic motion defines algebraically.

Figure 1 · The shadow of a rotating ball O P P′ view edge-on ↓ O −A +A P′ to and fro along the diameter
The reference particle $P$ moves uniformly on the circle. Its foot of perpendicular $P'$ on the horizontal diameter executes to-and-fro motion between $-A$ and $+A$ — this is the SHM. Viewed edge-on, $P$ and $P'$ coincide.

The reference circle, made precise

NCERT formalises the picture with Fig. 13.10. A particle $P$ moves uniformly on a circle of radius $A$ with angular speed $\omega$, the sense of rotation anticlockwise. At $t = 0$ the position vector $\vec{OP}$ makes an angle $\varphi$ with the positive $x$-axis. After time $t$ the radius has swept a further angle $\omega t$, so $\vec{OP}$ now makes the angle $(\omega t + \varphi)$ with the $x$-axis. Drop a perpendicular from $P$ to the $x$-axis; call the foot $P'$. The $x$-coordinate of $P'$ — its position on the diameter — is the cosine component of the radius:

$$x(t) = A\cos(\omega t + \varphi)$$

This is the defining equation of SHM. The particle $P$ and the circle it travels are called the reference particle and the reference circle, and they form the bridge between the rotation and the oscillation. Project instead onto the $y$-axis and the foot of the perpendicular gives $y(t) = A\sin(\omega t + \varphi)$ — again SHM, of the same amplitude $A$, but lagging the $x$-projection by a phase of $\pi/2$.

Figure 2 · The reference circle x y O P P′ A cos(ωt+φ) A sin(ωt+φ) ωt+φ A ω
The radius $\vec{OP}$ rotates with constant angular speed $\omega$. At time $t$ it makes angle $(\omega t + \varphi)$ with the $x$-axis. The $x$-projection $P'$ obeys $x = A\cos(\omega t + \varphi)$; the $y$-projection obeys $y = A\sin(\omega t + \varphi)$.

Why the projection is exactly SHM

The reference circle is more than an analogy; the two projections are SHMs because the geometry forces them to be. The reference particle's position vector has fixed length $A$ and rotates at fixed rate $\omega$. Resolving that constant-length, uniformly-turning vector onto a fixed axis gives a coordinate that is a pure sinusoid of $t$ — and a pure sinusoid of time is the algebraic definition of SHM. The NIOS account reaches the same conclusion from the dynamics: the centripetal acceleration of $P$ has magnitude $A\omega^2$, its component along the projection axis is $A\omega^2\sin\omega t$, and the force on the foot $P'$ comes out as $F = -m\omega^2 y$, a restoring force proportional to displacement. Either route lands on the same statement.

The circle-to-SHM dictionary

Every constant of the SHM equation reads off a feature of the reference circle. This is the single most useful table in the topic — it converts a rotation problem into an oscillation answer with no calculus.

Reference-circle quantityIts SHM projectionMeaning
Radius of the circle, $A$Amplitude $A$ in $x = A\cos(\omega t + \varphi)$Maximum displacement of the oscillation equals the radius
Angular speed $\omega$ of the radiusAngular frequency $\omega$, with $\omega = \dfrac{2\pi}{T} = 2\pi\nu$Rate of rotation equals the rate at which phase advances
Angle swept, $(\omega t + \varphi)$Phase of the motion, $(\omega t + \varphi)$Angular position of $P$ is the phase of $P'$
Initial angle $\varphi$ at $t = 0$Phase constant (initial phase) $\varphi$Starting angle of the radius fixes where the SHM begins
Tangential speed $v = \omega A$Velocity $v(t) = -\omega A\sin(\omega t + \varphi)$Projection of the tangential velocity onto the diameter
Centripetal acceleration $\omega^2 A$ (toward $O$)Acceleration $a(t) = -\omega^2 A\cos(\omega t + \varphi) = -\omega^2 x$Projection of the centre-directed acceleration
i
Build from here

The reference circle delivers $v(t)$ and $a(t)$ geometrically — the full treatment, including phase relations between $x$, $v$, $a$, lives in velocity and acceleration in SHM.

Why ω is the angular frequency

The reference circle answers a question that the bare equation $x = A\cos(\omega t + \varphi)$ leaves mysterious: why is $\omega$ called an angular frequency when the oscillation moves along a straight line and rotates through no angle at all? The answer is that $\omega$ is the angular speed of the rotating radius in the reference circle. The radius sweeps a full $2\pi$ radians in one period $T$, so $\omega = 2\pi / T$, and since the frequency of oscillation is $\nu = 1/T$, this gives $\omega = 2\pi\nu$. The SHM inherits its "angular" frequency from the rotation that casts it, even though the shadow itself never turns.

This also fixes the SI unit. Because $\omega$ is an angle (in radians) swept per unit time, its unit is radians per second — not hertz. NCERT is explicit that $\omega$ "is $2\pi$ times the frequency of oscillation." The reference circle makes the factor of $2\pi$ visible: it is the number of radians in one complete trip around the circle.

Projecting the velocity

The reference particle $P$ moves with speed $v = \omega A$ (angular speed times radius), directed along the tangent to the circle at $P$. To find the velocity of the shadow $P'$, project this tangential velocity onto the diameter. From the geometry of NCERT's Fig. 13.11, the projection comes out as

$$v(t) = -\omega A\sin(\omega t + \varphi)$$

The negative sign records that, for an anticlockwise rotation in the first quadrant, the shadow is moving in the $-x$ direction. The magnitude $\omega A$ is the velocity amplitude. The same expression follows by differentiating $x(t)$ once with respect to time — the reference circle simply gives it without calculus.

Figure 3 · Projecting velocity and centripetal acceleration x O P v = ωA a = ω²A P′ −ωA sin(ωt+φ) −ω²A cos(ωt+φ) = −ω²x
The reference particle's tangential velocity $\omega A$ and centripetal acceleration $\omega^2 A$ (pointing to the centre $O$) project onto the diameter to give the SHM's velocity $-\omega A\sin(\omega t+\varphi)$ and acceleration $-\omega^2 x$. Because $a$ always points to $O$, the projected acceleration always points to the mean position.

Projecting the acceleration

The reference particle in uniform circular motion has a centripetal acceleration of magnitude $v^2/A = \omega^2 A$, directed along $PO$ — from the particle toward the centre. Projecting this centre-directed acceleration onto the diameter, NCERT's Fig. 13.12 gives the acceleration of the shadow:

$$a(t) = -\omega^2 A\cos(\omega t + \varphi) = -\omega^2 x(t)$$

Two NEET-critical facts fall straight out of this projection. First, the acceleration is proportional to the displacement, $a = -\omega^2 x$ — the algebraic signature of SHM. Second, the minus sign means the acceleration is always directed toward the mean position: for $x > 0$, $a < 0$, and for $x < 0$, $a > 0$. This is inevitable in the picture, because the centripetal acceleration of $P$ always points to $O$, so its shadow always points to the centre of the diameter. The geometry explains the restoring character of SHM without writing a force law.

Worked example — NCERT Example 13.4

NCERT Example 13.4

A reference particle $P$ moves anticlockwise on a circle of radius $A$. At $t = 0$, $\vec{OP}$ makes an angle of $45^\circ = \pi/4$ with the positive $x$-axis, and the period of revolution is $T = 4\,\text{s}$. Obtain the SHM of the $x$-projection of $\vec{OP}$.

Read off the constants. Radius $\Rightarrow$ amplitude $A$. Period $\Rightarrow$ angular frequency $\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{4} = \dfrac{\pi}{2}\,\text{rad s}^{-1}$. Initial angle $\Rightarrow$ phase constant $\varphi = \pi/4$.

Write the angle swept. After time $t$ the radius has turned a further $\omega t = \dfrac{\pi}{2}t$, so it makes the angle $\left(\dfrac{\pi}{2}t + \dfrac{\pi}{4}\right)$ with the $x$-axis.

Project onto the $x$-axis. The $x$-coordinate of the foot $P'$ is the cosine of that angle times the radius:

$$x(t) = A\cos\!\left(\frac{\pi}{2}t + \frac{\pi}{4}\right)$$

Answer. An SHM of amplitude $A$, period $4\,\text{s}$, and initial phase $\pi/4$ — every constant read directly off the reference circle, no calculus required.

Quick recap

The circle of reference in one breath

  • SHM is the projection of uniform circular motion onto a diameter: $x = A\cos(\omega t + \varphi)$, $y = A\sin(\omega t + \varphi)$.
  • Radius of the reference circle $=$ amplitude $A$. Read it straight off.
  • Angle swept $(\omega t + \varphi)$ $=$ phase. Initial angle $=$ phase constant $\varphi$. Phase is an angle in radians, never a length.
  • $\omega$ is the angular speed of the rotating radius — hence "angular frequency," with $\omega = 2\pi/T = 2\pi\nu$ in rad s$^{-1}$.
  • Tangential velocity $\omega A$ projects to $v = -\omega A\sin(\omega t+\varphi)$; centripetal acceleration $\omega^2 A$ projects to $a = -\omega^2 x$, always toward the mean position.
  • The kinematics matches, but the SHM force $-kx$ is not the centripetal force — keep them separate.

NEET PYQ Snapshot — SHM and the Reference Circle

Three PYQs that lean on the reference-circle reading of SHM: read constants off the equation, and use $a = -\omega^2 x$.

NEET 2024

If $x = 5\sin(\pi t + \pi/3)$ m represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are:

  1. 5 cm, 2 s
  2. 5 m, 2 s
  3. 5 cm, 1 s
  4. 5 m, 1 s
Answer: (2) 5 m, 2 s

Reference-circle reading. The coefficient of the sine is the radius of the reference circle $=$ amplitude $= 5\,\text{m}$. The angular speed is $\omega = \pi\,\text{rad s}^{-1}$, so the period $T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{\pi} = 2\,\text{s}$. Amplitude and $\omega$ are read straight off the equation, exactly as off the circle.

NEET 2018

A pendulum moving freely to and fro like a simple harmonic oscillator has an acceleration of $20\ \text{m s}^{-2}$ at a distance of $5\ \text{m}$ from the mean position. The time period of oscillation is:

  1. $2\pi$ s
  2. $\pi$ s
  3. 2 s
  4. 1 s
Answer: (2) π s

Projected-acceleration law. The reference circle gives $a = -\omega^2 x$, so $|a| = \omega^2 x$. Then $20 = \omega^2(5) \Rightarrow \omega^2 = 4 \Rightarrow \omega = 2\,\text{rad s}^{-1}$. Hence $T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{2} = \pi\,\text{s}$.

NEET 2020

The phase difference between displacement and acceleration of a particle in simple harmonic motion is:

  1. $\dfrac{3\pi}{2}$ rad
  2. $\dfrac{\pi}{2}$ rad
  3. Zero
  4. $\pi$ rad
Answer: (4) π rad

From the projections. Displacement $x = A\sin(\omega t + \varphi)$; acceleration $a = -\omega^2 A\sin(\omega t + \varphi) = \omega^2 A\sin(\omega t + \varphi + \pi)$. The acceleration is the displacement projection turned through $\pi$ — they are always anti-phase, which is exactly why $a$ points back toward the mean position.

FAQs — SHM and Uniform Circular Motion

The reference-circle questions NEET aspirants most often slip on.

Is the projection of uniform circular motion always simple harmonic?
Yes, provided the projection is taken onto a diameter (a straight line through the centre) and the circular motion is uniform — constant angular speed ω. The foot of the perpendicular from the reference particle onto that diameter satisfies x(t) = A cos(ωt + φ), the defining equation of SHM. If the circular motion is non-uniform, or you project onto a line that does not pass through the centre, the result is not pure SHM.
What does the radius of the reference circle represent in the SHM?
The radius A of the reference circle is exactly the amplitude of the SHM. As the cosine swings between +1 and −1, the projection moves between +A and −A, so the maximum displacement equals the radius. This is a frequent NEET conversion: read the radius off the circle and that number is the amplitude of the oscillation.
Why is ω called the angular frequency?
Because in the reference-circle picture ω is literally the angular speed of the rotating radius — the rate at which the phase angle (ωt + φ) sweeps round the circle, in radians per second. The reference particle covers 2π radians in one period, so ω = 2π/T = 2πν. The name "angular frequency" records that ω is an angle swept per unit time, even though the projected SHM is a straight-line oscillation with no rotation of its own.
How does the velocity of the SHM follow from the reference circle?
The reference particle has speed v = ωA directed along the tangent to the circle. Projecting this tangential velocity onto the diameter gives the velocity of the SHM, v(t) = −ωA sin(ωt + φ). The maximum speed ωA occurs at the centre (mean position, where the tangent is parallel to the diameter) and the speed is zero at the extremes (where the tangent is perpendicular to the diameter).
Does the centripetal acceleration of the reference particle explain the SHM acceleration?
Yes. The reference particle has centripetal acceleration of magnitude ω²A directed from the particle toward the centre O. Its projection onto the diameter is a(t) = −ω²A cos(ωt + φ) = −ω²x(t). This is why the SHM acceleration is proportional to displacement and always directed toward the mean position — it is the shadow of an acceleration that always points to the centre of the circle.
Do projections onto the x-axis and y-axis give the same SHM?
They give SHMs of the same amplitude A and same angular frequency ω, but differing in phase by π/2. The x-projection is x = A cos(ωt + φ) and the y-projection is y = A sin(ωt + φ). NCERT notes that the two projections are SHMs "of the same amplitude as that of the projection on x-axis, but differing by a phase of π/2." Two perpendicular SHMs of equal amplitude and a π/2 phase difference combine back into uniform circular motion.
Related Topics
Oscillations — Parent chapter Full chapter map: periodic motion, SHM, energy, pendulum, resonance. Simple Harmonic Motion x = A cos(ωt + φ); amplitude, phase, angular frequency defined. Velocity & Acceleration in SHM v = −ωA sin(ωt+φ), a = −ω²x; phase relations between x, v, a. Force Law for SHM F = −kx with k = mω²; the restoring-force definition. Periodic & Oscillatory Motion Period, frequency; why circular motion is periodic but not oscillatory. Energy in SHM Kinetic and potential energy exchange; total energy ½kA².