MHT-CET Physics · Oscillations

SHM: Displacement, Velocity, Acceleration and Phase

In simple harmonic motion the acceleration is proportional to the displacement and directed towards the mean position, a = −ω²x; so x = A sin(ωt + α), the speed at displacement x is ω√(A² − x²), the extremes are ωA and ω²A, and the phase ωt + α says where in the cycle the particle is.

All Oscillations notes MHT-CET Physics strategy

Why this matters

44 PYQs, 7 HARD — the largest page in the chapter. Twenty-four use the velocity–displacement relation — the speed at a given x, the displacement at a given speed, the period, frequency or amplitude from two positions, and the distance between two positions (the HARD ones); thirteen are phase and time — the time to reach a point, the distance covered in successive seconds, the phase difference between two motions; seven are forces — a platform that must not lose its load, two restoring forces acting together, damping. Three cards.

Concept 1 of 3: Velocity, Acceleration and Displacement

Everything follows from v² = ω²(A² − x²) and a = −ω²x. At the mean position the speed is largest, ωA, and the acceleration zero; at the extremes the speed is zero and the acceleration ω²A. Two positions with known speeds give two equations: subtract them and A drops out, leaving ω; divide and ω drops out, leaving A. Given accelerations instead, use a = ω²x to turn them into positions. Maximum velocity and maximum acceleration together give both A = v²/a and ω = a/v.

Definition

$v = \omega\sqrt{A^2 - x^2}$ , $a = -\omega^2 x$ ; $v_{\max} = \omega A$ , $a_{\max} = \omega^2 A$ .
Two positions $(x_1, v_1)$ , $(x_2, v_2)$ : $\omega^2 = \dfrac{v_2^2 - v_1^2}{x_1^2 - x_2^2}$ , $A^2 = \dfrac{v_1^2x_2^2 - v_2^2x_1^2}{v_1^2 - v_2^2}$ .
Distance between two positions given speeds u, V and accelerations $a_1 < a_2$ : $\dfrac{u^2 - V^2}{a_1 + a_2}$ .
From the extremes: $A = \dfrac{v_{\max}^2}{a_{\max}}$ ; path length $2A = \dfrac{2v_{\max}^2}{a_{\max}}$ .
Speed $\tfrac{1}{2}v_{\max}$ at $x = \tfrac{\sqrt{3}}{2}A$ ; $\tfrac{1}{3}v_{\max}$ at $\tfrac{2\sqrt{2}}{3}A$ . Amplitude × 2 with period ÷ 3 ⇒ $v_{\max}$ × 6.

SHM velocity

v = \omega\sqrt{A^2 - x^2}, \qquad a = -\omega^2 x

Worked example

A particle in SHM has speeds 10 cm/s at 6 cm and 20 cm/s at 3 cm from the mean position. Find ω and the amplitude.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 2 · Q43Hard

Example 1 · Oscillations · SHM Kinematics — Displacement, Velocity, Phase, and Damping

A particle performs linear SHM. At a particular instant, velocity of the particle is

u

and acceleration is

\alpha

while at another instant, velocity is

v

and acceleration is

\beta

(

0 < \alpha < \beta

). The distance between the two positions is

Subtracting the accelerations instead of adding

The two positions are on the same side, so their distance is x₂ − x₁ = (u² − V²)/(a₁ + a₂). Dividing by a₁ − a₂ gives x₁ + x₂ instead.

Drill 23 more on velocity, acceleration and displacement

Concept 2 of 3: Phase, and Time to Reach a Point

Write x = A sin ωt if the particle starts at the mean position, x = A cos ωt if it starts at an extreme. The phase ωt then tells you where it is: from the mean position it reaches A/2 at ωt = π/6 (T/12), A/√2 at π/4 (T/8), and the extreme at π/2 (T/4). Each full oscillation adds 2π of phase. Velocity leads displacement by π/2 and acceleration is opposite to displacement, π out of phase — so force against time is the displacement graph turned upside down.

Definition

From the mean: $x = A\sin\omega t$ ; from an extreme: $x = A\cos\omega t$ ; $\omega = \dfrac{2\pi}{T}$ .
Mean → $\tfrac{A}{2}$ : $\tfrac{T}{12}$ ; → $\tfrac{A}{\sqrt{2}}$ : $\tfrac{T}{8}$ ; → A: $\tfrac{T}{4}$ .
Successive seconds with T = 8 s from the mean: first second covers $\tfrac{A}{\sqrt{2}}$ , second covers $A - \tfrac{A}{\sqrt{2}}$ — ratio $1 : (\sqrt{2} - 1)$ .
Phase gap between motions of periods $T_1, T_2$ after time t: $2\pi t\left(\tfrac{1}{T_1} - \tfrac{1}{T_2}\right)$ . Two oscillations ⇒ phase $4\pi$ .
Phase relations: v leads x by $\tfrac{\pi}{2}$ ; a and F are $\pi$ out of phase with x.
Pendulum released from θ: linear displacement $L\theta\cos\left(\sqrt{\tfrac{g}{L}}\,t\right)$ .

Phase

x = A\sin(\omega t + \alpha), \qquad \omega = \frac{2\pi}{T}

Worked example

SHM of period 6 s starts at the mean position. When does it first reach half the amplitude, and what is its phase then?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 22 April Shift II · Q24Moderate

Example 2 · Oscillations · SHM Kinematics — Displacement, Velocity, Phase, and Damping

A particle performing linear S.H.M. has period 8 seconds. At time

t = 0

, it is in the mean position. The ratio of the distances travelled by the particle in the

1^{\text{st~}}

and

2^{\text{nd~}}

second is

\left( \cos45^{\circ}= 1/\sqrt{2} \right)

Using cos when the motion starts at the mean

cos ωt starts at the extreme. A particle released from the mean position follows sin ωt — mixing them swaps T/12 with T/6.

Drill 12 more on phase, and time to reach a point

Concept 3 of 3: Restoring Forces, Detachment and Damping

A restoring force F = −kx gives ω = √(k/m); two such forces acting together add their k's, so 1/T² = 1/T₁² + 1/T₂². A load resting on a platform in vertical SHM stays on only while the platform's downward acceleration never exceeds g: ω²A ≤ g. A mass hung on a spring that is unstretched at the top of its swing has amplitude equal to the static stretch g/ω², and its top speed g/ω. Damping makes the amplitude die away exponentially — a factor each equal time — and slightly lowers ω.

Definition

$a = -bx$ ⇒ $\omega = \sqrt{b}$ , $T = \dfrac{2\pi}{\sqrt{b}}$ .
Two forces together: $T = \dfrac{T_1T_2}{\sqrt{T_1^2 + T_2^2}}$ .
Detachment: $\omega^2A = g$ ⇒ $A = \dfrac{gT^2}{4\pi^2}$ (T = 1 s ⇒ 0.25 m); least period for amplitude A: $2\pi\sqrt{\tfrac{A}{g}}$ .
Unstretched at the top: $A = \dfrac{g}{\omega^2}$ , $v_{\max} = \dfrac{g}{\omega}$ .
Damped: $A = A_0e^{-\lambda t}$ (⅓ in 2 s ⇒ 1/27 in 6 s); $\omega' = \sqrt{\dfrac{k}{m} - \left(\dfrac{b}{2m}\right)^2}$ .
Equal masses on springs $K_1, K_2$ with equal top speeds: $\dfrac{A_B}{A_A} = \sqrt{\dfrac{K_1}{K_2}}$ .

Damped amplitude

A = A_0e^{-\lambda t}, \qquad \omega' = \sqrt{\frac{k}{m} - \left(\frac{b}{2m}\right)^2}

Worked example

A platform oscillates vertically with period 2 s. Largest amplitude that keeps a coin on it (g = π² m/s²)?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 2 · Q37Moderate

Example 3 · Oscillations · SHM Kinematics — Displacement, Velocity, Phase, and Damping

A block of mass M rests on a piston executing S.H.M. of period one second. The amplitude of oscillations so that the mass is separated from the piston is (

g = 10\,\text{ms}^{-2}

\pi^2 = 10

)

Thinking damping only shrinks the amplitude

Damping also lowers the angular frequency: ω′ = √(k/m − b²/4m²). Adding the damping term, or leaving out the square root, gives the wrong options.

Drill 6 more on restoring forces, detachment and damping

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Velocity, Acceleration and Displacement
SHM velocity
$v = \omega\sqrt{A^2 - x^2}, \qquad a = -\omega^2 x$
Phase, and Time to Reach a Point
Phase
$x = A\sin(\omega t + \alpha), \qquad \omega = \frac{2\pi}{T}$
Restoring Forces, Detachment and Damping
Damped amplitude
$A = A_0e^{-\lambda t}, \qquad \omega' = \sqrt{\frac{k}{m} - \left(\frac{b}{2m}\right)^2}$

Watch out for (3)

Subtracting the accelerations instead of adding
→ Velocity, Acceleration and Displacement
Using cos when the motion starts at the mean
→ Phase, and Time to Reach a Point
Thinking damping only shrinks the amplitude
→ Restoring Forces, Detachment and Damping

Test yourself on Oscillations

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Oscillations chapter test Drill all 44 questions

Related notes

Energy in SHM →