Oscillations ISC Class 11: Simple Harmonic Motion Notes

What Oscillations is about

Weightage: Oscillations belongs to the unit Oscillations and Waves, which CISCE weights at 10 of the 70 marks in the ISC Class XI Physics Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Physics syllabus, examination year 2028, Class XI unit weightage

Key concepts

Periodic motion, period, frequency and periodic functions

Periodic motion repeats itself after equal intervals of time (the Earth's orbit, a clock's hands). Oscillatory (vibratory) motion is periodic motion to and fro about a fixed mean position (a pendulum, a loaded spring). Every oscillation is periodic, but not every periodic motion is an oscillation. Period T: the time for one complete oscillation (s). Frequency f = 1/T: oscillations per second (Hz). Angular frequency ω = 2πf = 2π/T (rad s⁻¹). Displacement in periodic motion is described by periodic functions: sin ωt and cos ωt repeat when ωt increases by 2π, so they have period 2π/ω. Any periodic function can be built from sines and cosines of different periods.

Worked example

The mains voltage alternates with a frequency of 50 Hz. Find its period and angular frequency.

Try it first, then show the solutionSolution

1. T = 1/f = 1/50 = 0.02 s
2. ω = 2πf = 2π × 50 = 314 rad s⁻¹

Answer: 0.02 s; 314 rad s⁻¹

Simple harmonic motion: definition, equation and phase

A body performs SHM when its acceleration (or restoring force) is directly proportional to its displacement from the mean position and is always directed towards it: F = −ky, a = −ω²y. From F = ma: $m\frac{d^{2}y}{d{t}^2}=-ky$, i.e. $\frac{d^{2}y}{d{t}^2}+{\omega }^{2}y=0$ with $\omega =\sqrt{\frac{k}{m}}$. Its general solution is $y=A\sin (\omega t+{\varphi }_0)$ (or a cosine form). Amplitude A: the maximum displacement. Phase (ωt + φ₀): the angle that fixes the position and direction of motion at time t. Initial phase (epoch) φ₀: the phase at t = 0. Two particles in SHM are in phase if their phase difference is 0, 2π…, and in opposite phase if it is π. Period $T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{k}}$, independent of amplitude.

Worked example

A particle moves according to y = 0.05 sin(10πt + π/6) m. Find its amplitude, frequency, period, initial phase and displacement at t = 0.

Try it first, then show the solutionSolution

1. A = 0.05 m; ω = 10π rad s⁻¹, so f = 5 Hz and T = 0.2 s
2. Initial phase = π/6
3. y(0) = 0.05 sin(π/6) = 0.025 m

Answer: 0.05 m; 5 Hz; 0.2 s; π/6; 2.5 cm

Velocity, acceleration and their graphs

Differentiating $y=A\sin (\omega t+{\varphi }_0)$: velocity $v=A\omega \cos (\omega t+{\varphi }_0)$, which can be written $v=\omega \sqrt{A^2-y^2}$; maximum $A\omega$ at the mean position, zero at the extremes. Acceleration $a=-A{\omega }^2\sin (\omega t+{\varphi }_0)=-{\omega }^{2}y$; maximum $A{\omega }^2$ at the extremes, zero at the mean position. Graphs (with φ₀ = 0): y-t is a sine curve; v-t is a cosine curve, leading y by π/2; a-t is a negative sine curve, leading v by π/2 and in opposite phase (π) to y. The v-y graph is an ellipse and the a-y graph a straight line of slope −ω² through the origin.

Worked example

A particle executes SHM with amplitude 0.1 m and period 2 s. Find its maximum speed, maximum acceleration, and speed when it is 0.06 m from the mean position.

Try it first, then show the solutionSolution

1. ω = 2π/2 = π rad s⁻¹; $v_{max}=A\omega =0.1\pi =0.314m{s}^{-1}$
2. $a_{max}=A{\omega }^2=0.1{\pi }^2=0.987m{s}^{-2}$
3. $v=\pi \sqrt{0.01-0.0036}=\pi \times 0.08=0.251m{s}^{-1}$

Answer: 0.314 m s⁻¹; 0.987 m s⁻²; 0.251 m s⁻¹

SHM as the projection of uniform circular motion

Let a point P move round a circle of radius A (the reference circle) with constant angular speed ω. Drop a perpendicular from P onto a diameter; its foot N moves to and fro along the diameter. If P starts at angle φ₀, then at time t the radius OP makes angle (ωt + φ₀) and N is at $y=A\sin (\omega t+{\varphi }_0)$: N performs SHM with amplitude equal to the radius, angular frequency equal to P's angular speed, and phase equal to the angle OP makes. Projecting P's velocity (Aω, along the tangent) gives $v=A\omega \cos (\omega t+{\varphi }_0)$, and projecting its centripetal acceleration (Aω², towards O) gives $a=-{\omega }^{2}y$. This picture makes the phase relations easy to see.

Worked example

A point on the rim of a wheel of radius 0.3 m rotating at 120 rpm is viewed edge-on, so its shadow moves along a straight line. Find the amplitude, maximum speed and maximum acceleration of the shadow.

Try it first, then show the solutionSolution

1. Amplitude = radius = 0.3 m; ω = 120 × 2π/60 = 4π rad s⁻¹
2. $v_{max}=A\omega =0.3\times 4\pi =3.77m{s}^{-1}$
3. $a_{max}=A{\omega }^2=0.3\times 16{\pi }^2=47.4m{s}^{-2}$

Answer: 0.3 m; 3.77 m s⁻¹; 47.4 m s⁻²

Energy in SHM

Potential energy at displacement y is the work done against the restoring force: $U={\int }_0^{y}kydy=\frac{1}{2}k{y}^2=\frac{1}{2}m{\omega }^2{y}^2$. Kinetic energy $K=\frac{1}{2}m{v}^2=\frac{1}{2}m{\omega }^2(A^2-y^2)$. Total energy $E=K+U=\frac{1}{2}m{\omega }^2{A}^2=\frac{1}{2}k{A}^2$, constant and proportional to the square of the amplitude. At the mean position E is all kinetic; at the extremes, all potential. Graphs against y: U is a parabola opening upward, K a parabola opening downward, and E a horizontal line. Against time, K and U each vary at twice the frequency of the motion.

Worked example

A 0.2 kg mass on a spring of constant 80 N m⁻¹ oscillates with amplitude 5 cm. Find the total energy and the displacement at which KE equals PE.

Try it first, then show the solutionSolution

1. E = ½kA² = ½ × 80 × 0.05² = 0.1 J
2. K = U means each is E/2: ½ky² = ¼kA², so y = A/√2
3. y = 0.05/1.414 = 0.0354 m

Answer: 0.1 J; about 3.5 cm from the mean position

Oscillations of a spring: horizontal and vertical

Horizontal spring: a mass m on a smooth surface attached to a spring of force constant k (restoring force per unit extension, N m⁻¹). Displaced by y, it feels F = −ky, so it performs SHM with $\omega =\sqrt{\frac{k}{m}}$ and $T=2\pi \sqrt{\frac{m}{k}}$. Vertical spring: hanging m stretches the spring by e until ke = mg (new equilibrium). A further displacement y gives a net force −k(e + y) + mg = −ky, so the motion is again SHM about the new equilibrium with the same $T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{e}{g}}$. Gravity shifts the mean position but does not change the period. T depends on m and k only, not on the amplitude or on g.

Worked example

A 0.5 kg mass hung from a spring stretches it by 9.8 cm. Find the spring constant and the period of vertical oscillations. (g = 9.8 m s⁻²)

Try it first, then show the solutionSolution

1. k = mg/e = 0.5 × 9.8/0.098 = 50 N m⁻¹
2. $T=2\pi \sqrt{\frac{0.5}{50}}=2\pi \times 0.1=0.628s$ (also $2\pi \sqrt{0.098/9.8}$)

Answer: 50 N m⁻¹; 0.63 s

The simple pendulum

A simple pendulum is a heavy point mass (bob) hanging from a rigid support by a light, inextensible string of length l. Displaced through angle θ, the forces on the bob are its weight mg and the tension T. The component mg cos θ balances the tension; the component mg sin θ acts along the arc towards the mean position. For small θ (in radian), sin θ ≈ θ = x/l, so the restoring force is $F=-\frac{mg}{l}x$: SHM with ${\omega }^2=\frac{g}{l}$. Time period: $T=2\pi \sqrt{\frac{l}{g}}$. It is independent of the mass of the bob and (for small amplitudes) of the amplitude. A seconds pendulum has T = 2 s, so l = g/π² ≈ 0.993 m. In a lift accelerating upward at a, the effective g becomes g + a (T decreases); in free fall the pendulum does not oscillate.

Worked example

Find the length of a seconds pendulum where g = 9.8 m s⁻², and its period on the Moon, where g is one-sixth as much.

Try it first, then show the solutionSolution

1. $T=2=2\pi \sqrt{\frac{l}{9.8}}$, so $l=\frac{9.8}{{\pi }^2}=0.993m$
2. On the Moon $T^{\prime }=2\sqrt{6}=4.9s$

Answer: 0.993 m; about 4.9 s

Physics in real life: Oscillations

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.

Topics in this chapter

• SHM: Velocity, Acceleration and Energy
• Spring-Mass System and Simple Pendulum

Oscillations formula sheet

Period and frequency

$f=\frac{1}{T},\omega =2\pi f=\frac{2\pi }{T}$

Hz and rad s⁻¹.

SHM condition and equation

$F=-ky,\frac{d^{2}y}{d{t}^2}+{\omega }^{2}y=0,{\omega }^2=\frac{k}{m}$

Restoring force ∝ displacement.

Displacement

$y=A\sin (\omega t+{\varphi }_0)$

Phase (ωt + φ₀); epoch φ₀.

Velocity

$v=A\omega \cos (\omega t+{\varphi }_0)=\omega \sqrt{A^2-y^2}$

Max Aω at y = 0.

Acceleration

$a=-{\omega }^{2}y$

Max Aω² at the extremes.

Energy

$U=\frac{1}{2}k{y}^2,K=\frac{1}{2}k(A^2-y^2),E=\frac{1}{2}k{A}^2=\frac{1}{2}m{\omega }^2{A}^2$

E constant.

Spring-mass system

$T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{e}{g}}(\text{vertical})$

e = static extension.

Simple pendulum

$T=2\pi \sqrt{\frac{l}{g}}$

Small amplitude; seconds pendulum l ≈ 0.993 m.

Oscillations: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Oscillations

The question patterns to expect, and the method that works for each.

Tricks to remember Oscillations

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

• Minus sign means SHM

  a = −ω²y: acceleration is always opposite to displacement, pulling back to the middle.

  Use it when: Checking whether a motion is SHM.

• Max speed in the middle, max pull at the ends

  Mean: v max, a zero. Extreme: v zero, a max.

  Use it when: Qualitative SHM questions.

• Quarter-cycle ladder

  y → v → a: each leads the previous by π/2 (so a is π ahead of y).

  Use it when: Phase-difference questions.

• Spring: mass on top; pendulum: length on top

  T = 2π√(m/k) and T = 2π√(l/g): the 'lazy' quantity on top makes it slower.

  Use it when: Recalling the two period formulae.

• Equal energies at A over root 2

  K = U at y = A/√2 ≈ 0.71A, not at A/2.

  Use it when: Energy-sharing questions.

Oscillations: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Oscillations

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

• ✗Mistake: Defining SHM without the direction

  Students write 'acceleration is proportional to displacement', leaving out 'directed towards the mean position'.

  Wrong working

  SHM: a ∝ y.

  Right working

  SHM: a ∝ −y; the acceleration is proportional to displacement and always directed towards the mean position.

  ✓Do this: Always include the minus sign or the direction in words.

  Usually costs: 1 mark.

• ✗Mistake: Putting K = U at y = A/2

  Students halve the amplitude instead of the energy.

  Wrong working

  Energies are equal halfway, at y = A/2.

  Right working

  ½ky² = ¼kA² gives y = A/√2 ≈ 0.71A.

  ✓Do this: Equate U to E/2 and solve.

  Usually costs: 1 mark.

• ✗Mistake: Using degrees in the small-angle step

  The approximation sin θ ≈ θ holds only with θ in radian.

  Wrong working

  sin 5° ≈ 5.

  Right working

  sin 5° ≈ 5π/180 = 0.087.

  ✓Do this: Write 'θ small, in radian' in every pendulum derivation.

  Usually costs: 1 mark.

• ✗Mistake: Thinking a vertical spring has a different period

  Students add g into the formula for a vertical spring.

  Wrong working

  Vertical spring: T = 2π√(m/(k + mg)).

  Right working

  Gravity only shifts the mean position; T = 2π√(m/k) = 2π√(e/g).

  ✓Do this: Measure the displacement from the new equilibrium.

  Usually costs: 1-2 marks.

• ✗Mistake: Reading ω as the frequency

  In y = A sin(4πt), students write f = 4π Hz.

  Wrong working

  f = 4π Hz, T = 1/(4π) s.

  Right working

  ω = 4π rad s⁻¹, f = ω/2π = 2 Hz, T = 0.5 s.

  ✓Do this: Divide ω by 2π for frequency.

  Usually costs: 1 mark.

Practice questions on Oscillations

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

1. Q1 · 1 mark · MCQExam level

   The period of a mass-spring system is proportional to:
   (a) m
   (b) √m
   (c) 1/m
   (d) m²

   Hint

   T = 2π√(m/k).

   Answer

   (b)

2. Q2 · 1 mark · MCQExam level

   The phase difference between the displacement and the acceleration of a particle in SHM is:
   (a) 0
   (b) π/4
   (c) π/2
   (d) π

   Hint

   a = −ω²y.

   Answer

   (d)

3. Q3 · 1 mark · Very short answerExam level

   Define simple harmonic motion.

   Hint

   Restoring force.

   Answer

   Periodic motion in which the restoring force (or acceleration) is proportional to the displacement from the mean position and always directed towards it.

4. Q4 · 1 mark · Very short answerExam level

   What is a seconds pendulum?

   Hint

   Period 2 s.

   Answer

   A simple pendulum with a period of 2 s (each swing one way takes 1 s); its length is about 0.993 m where g = 9.8 m s⁻².

5. Q5 · 2 marks · Short answerExam level

   Write expressions for the kinetic, potential and total energy of a particle in SHM at displacement y.

   Hint

   In terms of k and A.

   Answer

   K = ½k(A² − y²); U = ½ky²; E = ½kA² (constant).

6. Q6 · 2 marks · Short answerHarder exam level

   A 2 kg mass is attached to a spring of force constant 200 N m⁻¹. Find the period and frequency of its oscillations.

   Hint

   T = 2π√(m/k).

   Answer

   T = 2π√(0.01) = 0.628 s; f = 1.59 Hz.

7. Q7 · 2 marks · Short answerHarder exam level

   The displacement of a particle is y = 0.02 cos(50t + π/3) m. Find its frequency and its maximum velocity and acceleration.

   Hint

   ω = 50 rad s⁻¹ (not 50 Hz).

   Answer

   f = 50/2π = 7.96 Hz; v_max = Aω = 0.02 × 50 = 1.0 m s⁻¹; a_max = Aω² = 0.02 × 2500 = 50 m s⁻².

8. Q8 · 2 marks · Short answerHarder exam level

   At what displacement is the speed of a particle in SHM half its maximum value?

   Hint

   v = ω√(A² − y²).

   Answer

   A² − y² = A²/4, so y = (√3/2)A ≈ 0.87A.

9. Q9 · 2 marks · Short answerHarder exam level

   By what percentage does the period of a simple pendulum change if its length is increased by 21%?

   Hint

   T ∝ √l.

   Answer

   T' = T√1.21 = 1.1T: an increase of 10%.

10. Q10 · 3 marks · Short answerHarder exam level

    A 0.5 kg particle performs SHM of amplitude 4 cm and period π/5 s. Find its total energy and its KE and PE when it is 3 cm from the mean position.

    Hint

    ω = 2π/T = 10 rad s⁻¹.

    Answer

    E = ½ × 0.5 × 100 × 0.04² = 0.04 J; U = ½ × 0.5 × 100 × 0.03² = 0.0225 J; K = E − U = 0.0175 J.

11. Q11 · 3 marks · Short answerHarder exam level

    A coin sits on the rim of a record turntable of radius 0.15 m turning at 45 rpm. Its shadow on a wall moves in SHM. Find the amplitude, period, maximum speed and maximum acceleration of the shadow.

    Hint

    Reference circle: amplitude = radius, ω = angular speed.

    Answer

    A = 0.15 m; ω = 45 × 2π/60 = 4.71 rad s⁻¹, so T = 1.33 s; v_max = Aω = 0.707 m s⁻¹; a_max = Aω² = 3.33 m s⁻².

12. Q12 · 5 marks · Long answerHarder exam level

    (a) Draw graphs of displacement, velocity and acceleration against time for SHM starting at the mean position. [2]
    (b) Draw graphs of kinetic, potential and total energy against displacement. [2]
    (c) State the phase relation between velocity and displacement. [1]

    Hint

    Sine, cosine, negative sine; parabolas.

    Answer

    (a) y = A sin ωt; v = Aω cos ωt (starts at maximum); a = −Aω² sin ωt (mirror image of y). (b) U = ½ky² is an upward parabola touching E at ±A; K = ½k(A² − y²) is a downward parabola with maximum E at y = 0; E is a horizontal line. (c) Velocity leads displacement by π/2.

13. Q13 · 3 marks · Short answerHOTS / competency

    A pendulum clock keeps correct time at sea level. It is taken up a mountain where g is 0.1% smaller. Does it gain or lose time, and by how many seconds a day?

    Hint

    T ∝ g^(−1/2), so ΔT/T = −½ Δg/g.

    Answer

    T ∝ 1/√g, so ΔT/T = ½ × 0.1% = 0.0005: each swing takes longer and the clock loses time. Loss per day = 0.0005 × 86 400 = 43.2 s, about 43 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. 1.Differential equation of SHM?Show answer

   d²y/dt² + ω²y = 0.

2. 2.Velocity in SHM in terms of y?Show answer

   v = ω√(A² − y²).

3. 3.Maximum acceleration?Show answer

   Aω², at the extreme positions.

4. 4.Total energy of SHM?Show answer

   ½kA² = ½mω²A².

5. 5.Period of a mass on a spring?Show answer

   2π√(m/k).

6. 6.Period of a simple pendulum?Show answer

   2π√(l/g).

7. 7.What is the reference circle?Show answer

   A circle whose uniform motion projects onto a diameter as SHM.

8. 8.Length of a seconds pendulum?Show answer

   About 0.993 m (g = 9.8 m s⁻²).

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you define SHM and write its differential equation?
• ☐Can you read A, ω, T and the phase from y = A sin(ωt + φ)?
• ☐Can you find v and a at a given displacement?
• ☐Can you draw y-t, v-t and a-t graphs with correct phases?
• ☐Can you show SHM as the projection of uniform circular motion?
• ☐Can you find where KE equals PE?
• ☐Can you derive the period of a vertical spring-mass system?
• ☐Can you derive the period of a simple pendulum?

Oscillations: frequently asked questions