PhysicsISC Class 11 · Oscillations

Simple Harmonic Motion: Velocity, Acceleration, Phase and Energy

Updated 17 October 2026 · Fact-checked

In SHM the displacement is y = A sin(ωt + φ), the velocity is v = ω√(A² − y²), the acceleration is a = −ω²y and the total energy is ½mω²A² = ½kA², shared between kinetic and potential energy as the particle moves.

Understand SHM: Velocity, Acceleration and Energy

Definition. Simple harmonic motion is oscillatory motion in which the restoring acceleration is directly proportional to the displacement from the mean position and is directed towards it: a = −ω²y. Equivalent force law: F = −ky with ω² = k/m. The differential equation is d²y/dt² + ω²y = 0.

Kinematics. y = A sin(ωt + φ); v = dy/dt = Aω cos(ωt + φ) = ω√(A² − y²); a = −ω²y. The velocity is maximum (Aω) at the mean position, where a = 0; the acceleration is maximum (Aω²) at the extremes, where v = 0. T = 2π/ω and f = 1/T. The phase (ωt + φ) fixes the state of the motion; φ is the epoch (initial phase). The velocity leads the displacement by π/2 and the acceleration is in antiphase to the displacement.

Time between positions. Starting from the mean position with y = A sin ωt: y = A/2 at t = T/12, y = A/√2 at t = T/8, and y = A at t = T/4. The time to go from A/2 to A is T/4 − T/12 = T/6.

Reference circle. The projection of a particle moving uniformly in a circle of radius A and angular speed ω on a diameter performs SHM of amplitude A and angular frequency ω.

Energy. KE = ½mω²(A² − y²), PE = ½mω²y², total E = ½mω²A², constant (no damping). KE = PE at y = ±A/√2. The average KE and PE over a cycle are each E/2. E ∝ A² and E ∝ f².

SHM: Velocity, Acceleration and Energy: formulas and key facts

Displacement, velocity, acceleration: $y = A s in (ω t + φ) ∣ v = ω \sqrt (A^{2} - y^{2}) ∣ a = - ω^{2} y$; v_max = Aω; a_max = Aω².
Period and frequency: $T = 2 π / ω ∣ f = 1/ T$; ω = 2πf.
Energies: $K E = ½ m ω^{2} (A^{2} - y^{2}) ∣ P E = ½ m ω^{2} y^{2}$; Total E = ½mω²A².
Reference circle: $ω = an g u l a r s p ee d o n t h ec i r c l e ∣ A = r a d i u s$; Projection of uniform circular motion.

SHM: Velocity, Acceleration and Energy in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A block on a table vibrating

Suppose a 0.5 kg block on a smooth table oscillates in SHM with amplitude 0.04 m and period 0.8 s.

Find its total energy and its kinetic and potential energies at y = A/2.

Try it first, then show the solutionSolution

ω = 2π/0.8 = 7.854 rad s⁻¹; ω² = 61.69 s⁻².
E = ½mω²A² = 0.5 × 0.5 × 61.69 × (0.04)² = 0.0247 J.
At y = A/2: PE = E/4 = 0.0062 J; KE = 3E/4 = 0.0185 J.

Answer: E = 0.0247 J; KE = 0.0185 J and PE = 0.0062 J at y = A/2.

Memory hook: PE varies as y², so at half the amplitude it is a quarter of the total.

Tricks to remember SHM: Velocity, Acceleration and Energy

Short memory aids. Say each one aloud once, then use it on a question the same day.

Zero at the middle, maximum at the ends
At the mean position: v maximum, a zero. At the extremes: v zero, a maximum.
Use it when: MCQs on SHM quantities.
T/12, T/8, T/4
From the mean position, A/2, A/√2 and A are reached at T/12, T/8 and T/4.
Use it when: Time-between-positions questions.

SHM: Velocity, Acceleration and Energy: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Speed in SHM · 1 mark · MCQBoard level

In simple harmonic motion the speed of the particle is maximum:
(a) at the extreme positions
(b) at the mean position
(c) halfway between the mean and extreme positions
(d) at all positions

Try it first, then show the solutionSolution

v = ω√(A² − y²) is greatest when y = 0.

Answer: (b) at the mean position

Example 2: Reading an SHM equation · 3 marks · Short answerBoard level

The displacement of a particle is y = 0.05 sin(10πt) m. Find the amplitude, period, frequency, maximum speed and maximum acceleration.

Try it first, then show the solutionSolution

A = 0.05 m and ω = 10π rad s⁻¹, so T = 2π/ω = 0.2 s and f = 5 Hz. (1.5 marks)
v_max = Aω = 0.05 × 10π = 0.5π = 1.57 m s⁻¹. (0.75 mark)
a_max = Aω² = 0.05 × 100π² = 5π² = 49.3 m s⁻². (0.75 mark)

Answer: A = 5 cm; T = 0.2 s; f = 5 Hz; v_max = 1.57 m s⁻¹; a_max = 49.3 m s⁻².

Example 3: Velocity at a given displacement · 3 marks · Short answerHarder board level

A particle executes SHM with amplitude 10 cm and period 2 s. Find its speed and acceleration at a displacement of 6 cm from the mean position.

Try it first, then show the solutionSolution

ω = 2π/T = π rad s⁻¹. (1 mark)
v = ω√(A² − y²) = π√(100 − 36) = 8π = 25.1 cm s⁻¹. (1 mark)
a = ω²y = π² × 6 = 59.2 cm s⁻², towards the mean position. (1 mark)

Answer: v = 25.1 cm s⁻¹; a = 59.2 cm s⁻² towards the mean position.

Example 4: Time between positions · 4 marks · Short answerHarder board level

A particle in SHM has period 2.4 s and starts from the mean position. Find the time it takes to reach half of its amplitude for the first time, and the time to go from half amplitude to the extreme position.

Try it first, then show the solutionSolution

y = A sin ωt = A/2 → ωt = π/6. (1 mark)
t = (π/6)/(2π/T) = T/12 = 0.2 s. (1 mark)
Mean to extreme takes T/4 = 0.6 s. (1 mark)
Half amplitude to extreme = 0.6 − 0.2 = 0.4 s (= T/6). (1 mark)

Answer: 0.2 s; 0.4 s.

Example 5: Energy exchange · 5 marks · Long answerHOTS / competency

A 0.5 kg block executes SHM with amplitude 0.04 m and period 0.8 s. Find the total energy, the kinetic and potential energies at y = A/2, and the displacement at which the kinetic and potential energies are equal.

Try it first, then show the solutionSolution

ω = 2π/0.8 = 7.854 rad s⁻¹. (1 mark)
E = ½mω²A² = 0.5 × 0.5 × 61.69 × 1.6 × 10⁻³ = 0.0247 J. (1 mark)
At y = A/2: PE = ½mω²(A/2)² = E/4 = 6.2 × 10⁻³ J. (1 mark)
KE = E − PE = 1.85 × 10⁻² J. (1 mark)
KE = PE → ½mω²(A² − y²) = ½mω²y² → y = A/√2 = 0.0283 m. (1 mark)

Answer: E = 0.0247 J; KE = 0.0185 J, PE = 0.0062 J at y = A/2; KE = PE at y = 2.83 cm.

Where students lose marks in SHM: Velocity, Acceleration and Energy

The slips examiners see most often, and what to write instead.

Mistake: Using degrees in calculators
Evaluating sin(10πt) with the calculator in degree mode.
Do this: Use radian mode for sin(ωt).
Usually costs: 1 to 2 marks.
Mistake: Omitting the direction of acceleration
Quoting a = ω²y without the direction.
Do this: State that a is directed towards the mean position (a = −ω²y).
Usually costs: 1 mark.
Mistake: Mixing T and ω
Taking ω = T or using f instead of ω in v = ω√(A² − y²).
Do this: ω = 2π/T = 2πf.
Usually costs: 1 to 2 marks.

Exam tips

Compare the given equation with y = A sin(ωt + φ) and read A, ω and φ.
Define SHM using the condition a = −ω²y (towards the mean position).
Check energy answers with KE + PE = ½mω²A².
For time questions, use the reference circle and the phase angle.

Practice questions on SHM: Velocity, Acceleration and Energy

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The total energy of a particle in SHM is proportional to:
(a) A
(b) A²
(c) 1/A
(d) A³
Hint
E = ½mω²A².
Answer
(b) A²
Q2 · 2 marks · Short answerBoard level
A particle has displacement x = 4 cos(2πt) cm. Find the period, and the displacement at t = 0.25 s.
Hint
ω = 2π.
Answer
T = 2π/2π = 1 s. At t = 0.25 s, x = 4 cos(π/2) = 0.
Q3 · 3 marks · Short answerHarder board level
The acceleration of a particle in SHM is a = −36y (SI). Find the period and, for an amplitude of 0.1 m, the maximum speed.
Hint
a = −ω²y so ω² = 36.
Answer
ω = 6 rad s⁻¹, T = 2π/6 = 1.05 s. v_max = Aω = 0.1 × 6 = 0.6 m s⁻¹.
Q4 · 3 marks · Short answerHarder board level
Two SHMs y₁ = A sin ωt and y₂ = A sin(ωt + π/3) have the same period T. What is the phase difference in degrees and the time lag?
Hint
Phase difference ÷ ω gives the time lag.
Answer
Phase difference = π/3 = 60°. Time lag = (π/3)/ω = T/6.
Q5 · 3 marks · Short answerHarder board level
A particle in SHM has an amplitude of 8 cm. At what displacements is (a) its KE equal to its PE, and (b) its speed half its maximum?
Hint
Use KE = PE and v = ω√(A² − y²).
Answer
(a) y = A/√2 = 5.66 cm. (b) √(A² − y²) = A/2 → y = (√3/2)A = 6.93 cm.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a particle moves uniformly on a circle of radius 0.2 m with a period of 4 s, and its shadow on a diameter is observed. Take the shadow to be at the end of the diameter at t = 0.
(i) Find the maximum speed of the shadow. (1 mark)
(ii) Find its maximum acceleration. (1 mark)
(iii) Where is it at t = 1 s? (1 mark)
(iv) What is its speed at t = 1 s? (1 mark)
Hint
The shadow is SHM with A = 0.2 m and ω = 2π/4.
Answer
(i) ω = π/2; v_max = Aω = 0.314 m s⁻¹. (ii) a_max = Aω² = 0.2 × 2.467 = 0.493 m s⁻². (iii) y = A cos(ωt) = 0.2 cos(π/2) = 0, the mean position. (iv) The speed there is maximum: 0.314 m s⁻¹.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Differential equation of SHM?Show answer
d²y/dt² + ω²y = 0.
2.Velocity in terms of displacement?Show answer
v = ω√(A² − y²).
3.Where is KE equal to PE?Show answer
At y = A/√2.
4.Total energy of SHM?Show answer
½mω²A².

SHM: Velocity, Acceleration and Energy: frequently asked questions

Is every periodic motion SHM?

No. SHM requires the restoring acceleration proportional to the displacement, a = −ω²y. A ball bouncing on the floor is periodic but not SHM.

Why is the speed zero at the extreme positions?

The particle comes momentarily to rest there to reverse direction; all the energy is potential.

Does the period depend on the amplitude in SHM?

No. T = 2π/ω depends only on the system (m and k), not on A.