PhysicsCBSE Class 11 · Oscillations

SHM: Displacement, Velocity, Phase and Energy Class 11

Q: Is the motion of a particle on a circle SHM?

Its projection on a diameter is SHM, but the circular motion itself is not.

Q: What is the phase constant?

The value of the phase ωt + φ at t = 0; it fixes the starting position and direction.

Q: Why is the energy of SHM proportional to A²?

E = ½kA² and both K and U scale with the square of the displacement.

Q: Does the period depend on amplitude in SHM?

No; T = 2π/ω depends only on the system.

Q: Where is the speed maximum?

At the mean position, where the potential energy is zero.

Updated 17 October 2026 · Fact-checked

In simple harmonic motion x = A cos(ωt + φ), the velocity is v = −Aω sin(ωt + φ) with v² = ω²(A² − x²), and the acceleration is a = −ω²x. The speed is maximum (Aω) at the mean position and zero at the extremes, the acceleration is maximum (ω²A) at the extremes, and the total energy ½mω²A² stays constant, shifting between kinetic and potential.

Understand SHM: Displacement, Velocity, Phase and Energy

Definition. A particle performs simple harmonic motion (SHM) when its acceleration is proportional to its displacement from a fixed point and always directed towards that point: $\frac{d ^{2} x}{d t ^{2}} = - ω^{2} x$ . The solution is
$ $x = A cos (ω t + ϕ),$ $
with amplitude $A$ , angular frequency $ω = 2 π / T = 2 π f$ , and phase constant $ϕ$ (the phase is $ω t + ϕ$ ). SHM is the projection onto a diameter of uniform circular motion in a *reference circle* of radius $A$ and angular speed $ω$ .

Velocity and acceleration.
$ $v = - A ω sin (ω t + ϕ), a = - A ω^{2} cos (ω t + ϕ) = - ω^{2} x, v = \pm ω A^{2} - x^{2} .$ $
Maximum speed $A ω$ at $x = 0$ ; maximum acceleration $ω^{2} A$ at $x = \pm A$ . Velocity leads displacement by $π /2$ ; acceleration is $π$ out of phase with displacement.

**Finding $ϕ$ .** Use the initial position and the sign of the velocity. For example, with $x (0) = A /2$ and velocity towards the mean position (negative), $cos ϕ = \frac{1}{2}$ and $- A ω sin ϕ < 0$ , so $ϕ = π /3$ . Using two readings $(x_{1}, v_{1})$ and $(x_{2}, v_{2})$ in $v^{2} = ω^{2} (A^{2} - x^{2})$ gives $ω$ and $A$ .

Energy. $K = \frac{1}{2} m v^{2} = \frac{1}{2} m ω^{2} (A^{2} - x^{2})$ , $U = \frac{1}{2} m ω^{2} x^{2}$ (zero at the mean position), and
$ $E = K + U = \frac{1}{2} m ω^{2} A^{2} = constant .$ $
Both $K$ and $U$ oscillate at *twice* the frequency of the motion with average value $E /2$ . $K = U$ at $x = \pm A / 2$ . At $x = A /2$ , $K = \frac{3}{4} E$ and $U = \frac{1}{4} E$ . For a spring-mass system $m ω^{2} = k$ , so $E = \frac{1}{2} k A^{2}$ .

Time questions. From $x = A sin ω t$ (starting at the mean position), the time to reach $A /2$ is $T /12$ , and from $A /2$ to $A$ it is $T /6$ .

SHM: Displacement, Velocity, Phase and Energy: formulas and key facts

Displacement: $x = A cos (ω t + ϕ)$; $ω = 2 π / T$ .
Velocity: $v = \pm ω A^{2} - x^{2}, v_{ma x} = A ω$; At $x = 0$ .
Acceleration: $a = - ω^{2} x, a_{ma x} = ω^{2} A$; At the extremes.
Kinetic energy: $K = \frac{1}{2} m ω^{2} (A^{2} - x^{2})$; Zero at the extremes.
Potential energy: $U = \frac{1}{2} m ω^{2} x^{2} = \frac{1}{2} k x^{2}$; Zero at the mean position.
Total energy: $E = \frac{1}{2} m ω^{2} A^{2} = \frac{1}{2} k A^{2}$; Constant.

How to solve SHM: Displacement, Velocity, Phase and Energy questions

Extract A, ω and φ from the data first; then every other quantity is one substitution.

1Read A and ω (or T, f) from the equation or the data; convert units to SI.
2Use v² = ω²(A² − x²) and a = −ω²x for speed and acceleration at a given x.
3To find φ, apply the initial position and the sign of the velocity.
4For energy, use E = ½mω²A² and K = E − U.
5For times, use x = A sin ωt or A cos ωt and the reference circle angle.
6Check the signs of v and a with the direction of motion.

SHM: Displacement, Velocity, Phase 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 vibrating guitar string

Suppose a point on a string vibrates with amplitude 2.0 mm at 440 Hz as SHM.

Find the maximum speed and maximum acceleration of the point.

Try it first, then show the solutionSolution

ω = 2π × 440 = 2764 rad/s.
v_max = Aω = 2.0 × 10⁻³ × 2764 = 5.5 m/s; a_max = ω²A = 2764² × 2.0 × 10⁻³ = 1.5 × 10⁴ m/s².

Answer: 5.5 m/s; 1.5 × 10⁴ m/s².

Memory hook: The acceleration is thousands of times g.

A piston

Suppose the piston of a small engine moves with a stroke of 8.0 cm (twice the amplitude) at 50 rev/s as in SHM.

Find the maximum speed of the piston.

Try it first, then show the solutionSolution

A = 4.0 cm = 0.040 m; ω = 2π × 50 = 314 rad/s.
v_max = 0.040 × 314 = 12.6 m/s.

Answer: About 12.6 m/s.

Memory hook: The stroke is twice the amplitude.

Tricks to remember SHM: Displacement, Velocity, Phase and Energy

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

Speed max at the centre
Velocity is maximum at x = 0 and zero at x = ±A; acceleration is the opposite.
Use it when: Quick extremes.
K = ¾E at A/2
At x = A/2, K : U = 3 : 1.
Use it when: Fraction questions.
Double frequency
K and U oscillate at 2f.
Use it when: Energy graph.
Two readings give ω and A
Subtract the two v² equations.
Use it when: Data with two points.

SHM: Displacement, Velocity, Phase and Energy: solved examples

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

Example 1: Reading an SHM equation · 2 marks · Very short answerBoard level

A particle moves as x = 0.04 sin(10πt) metres. Find its amplitude, angular frequency, frequency and period.

Try it first, then show the solutionSolution

A = 0.04 m = 4 cm; ω = 10π rad/s. (1)
f = ω/2π = 5 Hz; T = 1/f = 0.2 s. (1)

Answer: 4 cm; 10π rad/s; 5 Hz; 0.2 s.

Example 2: Speed at a given displacement · 3 marks · Short answerBoard level

A particle in SHM has amplitude 5 cm and angular frequency 6 rad/s. Find its speed at a displacement of 3 cm, its maximum speed and its maximum acceleration.

Try it first, then show the solutionSolution

v = ω√(A² − x²) = 6 × √(25 − 9) = 6 × 4 = 24 cm/s. (1)
v_max = Aω = 5 × 6 = 30 cm/s. (1)
a_max = ω²A = 36 × 5 = 180 cm/s². (1)

Answer: 24 cm/s; 30 cm/s; 180 cm/s².

Example 3: Finding the phase constant · 3 marks · Short answerHarder board level

At t = 0 a particle in SHM of amplitude 6 cm and period 2 s is at x = +3 cm and moving towards the mean position. Write x(t) as A cos(ωt + φ) and find the position at t = 1 s and the first time it passes the mean position.

Try it first, then show the solutionSolution

ω = 2π/T = π rad/s. At t = 0, cos φ = 1/2; moving towards x = 0 means velocity negative, −Aω sin φ < 0, so φ = π/3. (1)
x = 6 cos(πt + π/3) cm; at t = 1 s: 6 cos(4π/3) = −3 cm. (1)
Mean position: πt + π/3 = π/2, so t = 1/6 s. (1)

Answer: x = 6 cos(πt + π/3) cm; −3 cm; 1/6 s.

Example 4: Energy exchange · 4 marks · Long answerHarder board level

A 0.40 kg block on a smooth horizontal spring of constant 250 N/m oscillates with amplitude 8.0 cm. Find (a) the total energy, (b) the kinetic and potential energies and the speed at x = 4.0 cm, (c) the maximum speed, and (d) the position at which K = U.

Try it first, then show the solutionSolution

(a) E = ½kA² = ½ × 250 × (0.08)² = 0.80 J. (1)
(b) U = ½ × 250 × (0.04)² = 0.20 J, K = 0.60 J, v = √(2 × 0.60/0.40) = 1.73 m/s. (1)
(c) ω = √(k/m) = 25 rad/s; v_max = Aω = 2.0 m/s. (1)
(d) K = U gives x = A/√2 = 5.7 cm. (1)

Answer: (a) 0.80 J; (b) 0.20 J, 0.60 J, 1.73 m/s; (c) 2.0 m/s; (d) 5.7 cm.

Example 5: Energy graph data · 4 marks · Long answerHOTS / competency

The kinetic energy of a particle in SHM varies with displacement as K = 0.18 − 50x² (joules, x in metres). Find (a) the amplitude, (b) the total energy, (c) the spring constant of the restoring force, (d) the mass if ω = 10 rad/s, and (e) U and K at x = 0.04 m.

Try it first, then show the solutionSolution

(a) K = 0 at the extreme: 50A² = 0.18, A² = 0.0036, A = 0.06 m. (1)
(b) E = K at x = 0 = 0.18 J. (1)
(c) U = E − K = 50x² = ½kx², so k = 100 N/m. (1)
(d) m = k/ω² = 100/100 = 1.0 kg. (e) U = 50 × 0.0016 = 0.08 J; K = 0.18 − 0.08 = 0.10 J. (1)

Answer: A = 6 cm; E = 0.18 J; k = 100 N/m; m = 1.0 kg; U = 0.08 J, K = 0.10 J.

Where students lose marks in SHM: Displacement, Velocity, Phase and Energy

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

Mistake: Using degrees on the calculator for sin or cos of ωt.
Mode error.
Do this: Use radians for ωt + φ.
Usually costs: The answer mark.
Mistake: Writing v_max = ω²A or a_max = ωA.
Mixing the formulas.
Do this: v_max = Aω, a_max = Aω².
Usually costs: Whole numerical.
Mistake: Ignoring the sign of velocity when finding φ.
Choosing φ from x(0) alone.
Do this: Use the direction of motion to pick between ±φ.
Usually costs: 1 mark.
Mistake: Saying total energy changes with x.
Confusing K, U and E.
Do this: E = ½kA² is constant; only K and U change.
Usually costs: Concept mark.

Exam tips

Write x = A cos(ωt + φ) and list A, ω, φ before anything else.
Mention the radian mode for trigonometric values.
For energy answers show E, then U, then K = E − U.
Draw the reference circle for time and phase questions.
Give units (cm vs m) consistently.

Practice questions on SHM: Displacement, Velocity, Phase 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 · MCQHarder board level
At a displacement of half the amplitude, the ratio of kinetic energy to potential energy of a particle in SHM is
(a) 1 : 3 (b) 3 : 1 (c) 1 : 4 (d) 4 : 1
Hint
K = ¾E, U = ¼E.
Answer
(b) 3 : 1.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The acceleration of a particle in SHM is always directed opposite to its displacement.
Reason (R): The acceleration is given by a = −ω²x.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Look at the sign.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
A body performs SHM with period 0.5 s and amplitude 2 cm. Find its maximum speed and maximum acceleration.
Hint
ω = 2π/T.
Answer
ω = 4π rad/s; v_max = 2 × 4π = 25.1 cm/s; a_max = ω²A = 16π² × 2 = 316 cm/s².
Q4 · 3 marks · Short answerHarder board level
A particle in SHM has a speed of 8 cm/s at a displacement of 3 cm and a speed of 6 cm/s at a displacement of 4 cm. Find ω and the amplitude.
Hint
Use v² = ω²(A² − x²) twice.
Answer
64 = ω²(A² − 9) and 36 = ω²(A² − 16). Subtracting: 28 = 7ω², ω = 2 rad/s. A² = 9 + 16 = 25, A = 5 cm.
Q5 · 4 marks · Long answerHOTS / competency
A point on a wheel of radius 0.20 m rotating at 3 rev/s casts a shadow on a wall that moves in SHM. Find the amplitude, period, maximum speed and maximum acceleration of the shadow, and the time the shadow takes to move from the extreme position to half the amplitude.
Hint
The reference circle has radius A and ω = 2π × 3.
Answer
A = 0.20 m, ω = 6π = 18.85 rad/s, T = 1/3 s. v_max = Aω = 3.77 m/s; a_max = ω²A = 71 m/s². From x = A cos ωt = A/2: ωt = π/3, so t = T/6 = 0.056 s.

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.v in terms of x?Show answer
ω√(A² − x²).
2.Maximum acceleration?Show answer
ω²A.
3.Total energy of SHM?Show answer
½mω²A².
4.K = U at what x?Show answer
A/√2.
5.Phase difference between a and x?Show answer
π.
6.Frequency of the energy oscillation?Show answer
2f.

SHM: Displacement, Velocity, Phase and Energy: frequently asked questions

Is the motion of a particle on a circle SHM?

Its projection on a diameter is SHM, but the circular motion itself is not.

What is the phase constant?

The value of the phase ωt + φ at t = 0; it fixes the starting position and direction.

Why is the energy of SHM proportional to A²?

E = ½kA² and both K and U scale with the square of the displacement.

Does the period depend on amplitude in SHM?

No; T = 2π/ω depends only on the system.

Where is the speed maximum?

At the mean position, where the potential energy is zero.