Lesson 5: Oscillations | State Board · Class 12 · Physics

Periodic motion and simple harmonic motion

From daily life

A swing pulled away from the middle is drawn back towards it, again and again.

Simple definition

Simple harmonic motion (SHM) is an oscillation in which the acceleration is proportional to the displacement and opposite to it.

In the figure : The shadow on the x-axis of a point moving on a circle performs SHM along the line.

Live figure

Periodic motion and simple harmonic motion

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Rule 1: Definition of SHM — A force proportional to the displacement and opposite to it : F = −k x ⇒ a = −ω² x (ω² = k ÷ m). This is the defining property of SHM.

Easy : Given a = −16 x ⇒ ω² = 16, ω = 4 rad/s.
Medium : The period of the same particle T = 2π ÷ 4 = 1.571 s.
Exam (CET) : a = −9 x ⇒ period? (1) 2π/3 s (2) 2π/9 s (3) 3 s (4) 9 s → answer 1). Trick : ω = √(the number), then 2π ÷ ω.

Rule 2: Periodic, oscillatory and SHM — Every SHM is periodic but not every periodic motion is SHM (e.g. the Earth's revolution — F is not ∝ x). SHM needs F ∝ −x.

Easy : A swing, a pendulum, a mass on a spring : SHM (for small oscillations).
Medium : A minute hand : periodic but not SHM.
Exam (CET) : If F ∝ −x² the motion is not SHM. (Options : 1) SHM 2) not SHM 3) resonance 4) damped → answer 2)

Rule 3: SHM and uniform circular motion — The projection of a particle in UCM on a diameter performs SHM : x = A sin(ωt + φ). The radius A = amplitude, the angular speed ω = angular frequency of the SHM, φ = initial phase.

Easy : Radius 0.2 m, 3 rev/s : A = 0.2 m, ω = 6π rad/s.
Medium : At t = 0, x = 0 and increasing ⇒ φ = 0.
Exam (CET) : One period = 2π turned on the circle ⇒ T = 2π ÷ ω. (Options : 1) π/ω 2) 2π/ω 3) ω/2π 4) 2πω → answer 2)

Derivation : how x = A sin(ωt + φ) arises

Step 1 : F = −k x and F = m d²x/dt² ⇒ d²x/dt² = −(k ÷ m) x = −ω² x.
Step 2 : Which function gives −ω² times itself on differentiating twice? sin or cos.
Step 3 : Check : x = A sin(ωt + φ) ⇒ d²x/dt² = −ω² A sin(ωt + φ) = −ω² x ✓.
Step 4 : A and φ are fixed by the initial conditions; ω = √(k ÷ m).

Remember

F = −kx; a = −ω²x; x = A sin(ωt + φ); T = 2π ÷ ω.

Common mistake / CET trap

ω (angular frequency) and n (frequency) differ : ω = 2πn. The circle of SHM is an aid — the real particle does not move on a circle.

Shortcut

Memory clue : "a equals −ω² x" – the minus sign and the ω².

Spot the difference

Quantity	Symbol	Unit / dimensions
Amplitude	A	m; [L]
Angular frequency	ω	rad/s; [T⁻¹]
Period	T = 2π/ω	s; [T]
Frequency	n = 1/T	Hz; [T⁻¹]
Force constant	k	N/m; [M T⁻²]

Practice questions

Easy

1State the condition for SHM.
2a = −25 x : ω?
3n = 5 Hz : T?

Medium

4a = −4 x : period?
5k = 100 N/m, m = 1 kg : ω?
6Period 0.5 s : ω?

Challenging

7UCM radius 5 cm, 2 rev/s : A and ω of the SHM?
8Why is F ∝ −x³ not SHM?

Show answers (try them yourself first)

1F ∝ −x (a = −ω² x)
2ω = 5 rad/s
3T = 0.2 s
4ω = 2, T = π = 3.14 s
5ω = √100 = 10 rad/s
6ω = 2π ÷ 0.5 = 12.57 rad/s
7A = 5 cm, ω = 4π = 12.57 rad/s
8The force is not proportional to x, so a is not ∝ −x

Equations and graphs of SHM

From daily life

A swing is fastest as it passes the middle and stops for a moment at the ends.

Simple definition

x, v, a are sine–cosine curves; their phase relations solve most questions.

In the figure : x (blue), v (green, π/2 ahead), a (red, opposite to x).

Rule 1: v and a formulas — x = A sin ωt ⇒ v = A ω cos ωt = ω √(A² − x²), a = −ω² x. Maxima : v_max = A ω (at the mean position), a_max = A ω² (at the extremes).

Easy : A = 0.1 m, T = 2 s : ω = π; v_max = 0.314 m/s.
Medium : For the same particle a_max = 0.1 × π² = 0.987 m/s².
Exam (CET) : A = 0.1 m, ω = 5 : at x = 0.06 m v = 5 × √(0.01 − 0.0036) = 0.4 m/s. (Options : 1) 0.4 2) 0.5 3) 0.3 4) 0.8 → answer 1). Trick : 6-8-10 triplet ⇒ √(A² − x²) = 0.08.

Rule 2: Phase relations — v leads x by π/2; a is in antiphase with x (π). At x = 0, v is maximum and a = 0; at x = ±A, v = 0 and a is maximum.

Easy : At x = A : v = 0, a = −ω² A.
Medium : At x = 0 : v = ± A ω, a = 0.
Exam (CET) : v_max ÷ a_max = 1 ÷ ω. (Options : 1) ω 2) 1/ω 3) ω² 4) 1/ω² → answer 2)

Rule 3: Initial phase and comparing two SHMs — Phase θ = ωt + φ. The phase difference Δφ of two SHMs : Δφ = 0 ⇒ in phase, Δφ = π ⇒ in antiphase.

Easy : x₁ = sin ωt and x₂ = sin(ωt + π) : in antiphase.
Medium : sin and cos differ by π/2.
Exam (CET) : If the ω's differ, the phase difference changes with time. (Options : 1) constant 2) changes 3) zero 4) π → answer 2)

Derivation : v = ω √(A² − x²)

Step 1 : x = A sin θ and v = dx/dt = A ω cos θ.
Step 2 : cos θ = √(1 − sin² θ) = √(1 − x² ÷ A²).
Step 3 : v = A ω √(1 − x² ÷ A²) = ω √(A² − x²).
Step 4 : a = dv/dt = −A ω² sin θ = −ω² x.

Remember

v = ω√(A² − x²); v_max = Aω; a_max = Aω²; v leads x by π/2.

Common mistake / CET trap

v_max and a_max occur at different places. Do not mix ω and n; ω = 2πn. Read the phase carefully in graphs.

Shortcut

Memory clue : "speed in the middle, acceleration at the ends".

Spot the difference

Quantity	Formula	Maximum where
x	A sin(ωt + φ)	extreme : A
v	ω √(A² − x²)	x = 0 : Aω
a	−ω² x	extreme : Aω²

Practice questions

Easy

1Formula for v_max?
2a at x = 0?
3v at x = A?

Medium

4A = 0.2 m, ω = 3 : v_max?
5A = 0.2 m, ω = 3 : a_max?
6A = 5 cm, ω = 10, x = 3 cm : v?

Challenging

7v_max = 4 m/s, a_max = 8 m/s² : ω and A?
8x = 0.1 sin(20t) : T?

Show answers (try them yourself first)

1A ω
2Zero
3Zero
40.6 m/s
51.8 m/s²
6v = 10 × 4 = 40 cm/s
7ω = 8 ÷ 4 = 2 rad/s; A = 4 ÷ 2 = 2 m
8T = 2π ÷ 20 = 0.314 s

Energy in simple harmonic motion

From daily life

On a swing the KE is high at the middle and the PE is high at the ends; the total stays the same.

Simple definition

In SHM energy keeps changing between KE and PE; the total mechanical energy E is constant (without friction).

In the figure : PE is a parabola, KE an inverted one; their sum is the horizontal line E.

Rule 1: KE, PE and total E — KE = ½ m ω² (A² − x²), PE = ½ m ω² x², E = ½ m ω² A² = ½ k A² (constant). All KE at the mean position, all PE at the extremes.

Easy : m = 0.2 kg, A = 0.05 m, ω = 10 : E = ½ × 0.2 × 100 × 0.0025 = 0.025 J.
Medium : At x = 0.03 m PE = ½ × 0.2 × 100 × 0.0009 = 0.009 J; KE = 0.016 J.
Exam (CET) : Amplitude doubled ⇒ E becomes? (1) 2 (2) 4 (3) 8 (4) 1/2 times → answer 2). Trick : E ∝ A².

Rule 2: Where KE = PE — ½ m ω² (A² − x²) = ½ m ω² x² ⇒ x = ± A ÷ √2 ≈ ± 0.707 A. There KE = PE = E ÷ 2.

Easy : A = 10 cm : KE = PE at x = 7.07 cm.
Medium : At x = A ÷ 2, KE : PE = (1 − ¼) : ¼ = 3 : 1.
Exam (CET) : At x = A ÷ 2, PE is what fraction of E? (1) 1/2 (2) 1/4 (3) 3/4 (4) 1/8 → answer 2)

Rule 3: Average energy and frequency — Over one period ⟨KE⟩ = ⟨PE⟩ = E ÷ 2. KE and PE oscillate at twice the SHM frequency (2n); E is constant.

Easy : n = 4 Hz ⇒ KE oscillates at 8 Hz.
Medium : E = 0.1 J ⇒ ⟨KE⟩ = 0.05 J.
Exam (CET) : In an SHM of T = 2 s, the period of the KE? (1) 2 s (2) 1 s (3) 4 s (4) 0.5 s → answer 2)

Derivation : E = ½ m ω² A²

Step 1 : Force F = −k x; PE = work against the force = ∫₀ˣ k x dx = ½ k x².
Step 2 : k = m ω² ⇒ PE = ½ m ω² x².
Step 3 : KE = ½ m v² = ½ m ω² (A² − x²).
Step 4 : Sum : E = KE + PE = ½ m ω² A² (x cancels).

Remember

E = ½ m ω² A²; KE = ½mω²(A² − x²); PE = ½mω²x²; KE = PE ⇒ x = A/√2.

Common mistake / CET trap

E ∝ A², not A. KE and PE oscillate at 2n. Do not put the amplitude A in place of x.

Shortcut

Memory clue : "E goes with A squared".

Spot the difference

Energy	Formula	Unit
KE	½ m ω² (A² − x²)	J; [M L² T⁻²]
PE	½ m ω² x²	J
E	½ m ω² A² = ½ k A²	J

Practice questions

Easy

1Formula for E?
2PE at the mean position?
3KE at the extremes?

Medium

4m = 1 kg, ω = 2, A = 0.1 m : E?
5k = 200 N/m, A = 0.05 m : E?
6E = 8 J, KE at x = A/2?

Challenging

7Where is KE = PE? A = 20 cm.
8Amplitude ×3 ⇒ E?

Show answers (try them yourself first)

1½ m ω² A²
2Zero
3Zero
4E = ½ × 1 × 4 × 0.01 = 0.02 J
5E = ½ × 200 × 0.0025 = 0.25 J
6KE = 8 × ¾ = 6 J
7x = 20 ÷ √2 = 14.1 cm
8E becomes 9 times

Spring–mass oscillator

From daily life

A car's suspension spring bobs up and down after a bump.

Simple definition

A mass on a spring, once stretched or compressed, performs SHM because of F = −kx.

In the figure : Horizontal on the left, vertical on the right : T = 2π√(m ÷ k) in both.

Rule 1: Time period — F = −k x ⇒ ω = √(k ÷ m) ⇒ T = 2π √(m ÷ k). T does not depend on the amplitude or g; larger m ⇒ larger T, larger k ⇒ smaller T.

Easy : m = 0.5 kg, k = 50 N/m : T = 2π √0.01 = 0.628 s.
Medium : m made four times : T doubles.
Exam (CET) : k = 400 N/m, T = 0.314 s ⇒ m = ? (1) 1 kg (2) 0.1 kg (3) 4 kg (4) 0.4 kg → answer 1). Trick : 0.314 = 2π ÷ 20 ⇒ ω = 20, m = k ÷ ω² = 1.

Rule 2: Combinations of springs — Parallel : k = k₁ + k₂ (smaller T). Series : 1 ÷ k = 1 ÷ k₁ + 1 ÷ k₂ (larger T). A spring cut in half ⇒ k of each half doubles (k ∝ 1 ÷ length).

Easy : k₁ = k₂ = 100 : parallel 200; series 50 N/m.
Medium : T(series) ÷ T(parallel) = √(200 ÷ 50) = 2.
Exam (CET) : A spring is cut in half : factor on T? (1) ÷√2 (2) ×√2 (3) ×2 (4) ÷2 → answer 1). Trick : k × 2 ⇒ T ÷ √2.

Rule 3: Vertical spring — Hanging a mass stretches the spring by Δl = m g ÷ k; SHM starts about the new equilibrium. T = 2π √(Δl ÷ g) = 2π √(m ÷ k) — g has no separate effect.

Easy : m = 0.2 kg, k = 20 : Δl = 2 ÷ 20 = 0.1 m; T = 2π √(0.1 ÷ 10) = 0.628 s.
Medium : The same oscillator on the Moon : same T (same m, k).
Exam (CET) : Δl = 4 cm ⇒ T = ? (g = 10) (1) 0.4 s (2) 0.126 s (3) 1.26 s (4) 0.63 s → answer 1). Check : 2π √(0.04 ÷ 10) = 0.397 s.

Derivation : T = 2π √(m ÷ k)

Step 1 : Displace the mass by x; the spring force is F = −k x.
Step 2 : Newton : m d²x/dt² = −k x ⇒ d²x/dt² = −(k ÷ m) x.
Step 3 : Compare with SHM (a = −ω² x) ⇒ ω² = k ÷ m.
Step 4 : T = 2π ÷ ω = 2π √(m ÷ k).

Remember

T = 2π√(m/k); parallel k₁ + k₂; series k₁k₂/(k₁ + k₂); half a spring has twice k.

Common mistake / CET trap

The T of the oscillator does not depend on g. Do not swap the series and parallel formulas.

Shortcut

Memory clue : "Big m = slow; big k = fast".

Spot the difference

Combination	k_eq	T
Parallel	k₁ + k₂	smaller
Series	k₁k₂ ÷ (k₁ + k₂)	larger
Cut in half	2k	T ÷ √2

Practice questions

Easy

1Formula for T?
2k = 80, m = 0.2 : ω?
3k_eq in parallel?

Medium

4m = 2 kg, k = 200 : T?
5k₁ = 30, k₂ = 60 in series : k_eq?
6m ×9 ⇒ T?

Challenging

7A 0.5 kg mass stretches a spring by 5 cm : T? (g = 10)
8Does T change on the Moon?

Show answers (try them yourself first)

12π √(m ÷ k)
2ω = √400 = 20 rad/s
3k₁ + k₂
4T = 2π √0.01 = 0.628 s
5k_eq = 1800 ÷ 90 = 20 N/m
6T triples
7Δl = 0.05 ⇒ T = 2π √(0.05 ÷ 10) = 0.444 s
8No (same m and k)

The simple pendulum

From daily life

A clock pendulum swings with a regular tick-tock.

Simple definition

A simple pendulum is a point bob on a weightless string; for small angles its motion is SHM.

In the figure : mg sin θ pulls it back; mg cos θ balances the string tension.

Live figure

The simple pendulum

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Rule 1: Time period — For small angles sin θ ≈ θ : T = 2π √(L ÷ g). T does not depend on m or amplitude; L ∝ T²; smaller g ⇒ larger T.

Easy : L = 1 m, g = 9.8 : T = 2.007 s.
Medium : Seconds pendulum (T = 2 s) : L = g ÷ π² = 0.993 m ≈ 1 m.
Exam (CET) : L made four times ⇒ T? (1) ×2 (2) ×4 (3) ×½ (4) ×16 → answer 1). Trick : T ∝ √L.

Rule 2: Pendulum in a lift / accelerated frame — Effective g' : lift accelerating up by a ⇒ g' = g + a; down ⇒ g − a; in free fall g' = 0 ⇒ T infinite. T = 2π √(L ÷ g').

Easy : g = 10, a = 2.5 upward : g' = 12.5; T/T₀ = √(10 ÷ 12.5) = 0.894.
Medium : g' = g − a for a downward acceleration : T larger.
Exam (CET) : Does a pendulum oscillate in a freely falling lift? (1) yes (2) no (3) T = 2π√(L/g) (4) T = 0 → answer 2)

Rule 3: Practical limits — Beyond about 10° the period increases slightly. The true length L = string length + radius of the bob (up to the centre of mass).

Easy : String 0.95 m, bob radius 0.05 m : L = 1.00 m.
Medium : L increases (metal rod in summer) ⇒ the clock runs slow.
Exam (CET) : On a mountain g is smaller ⇒ the clock? (1) fast (2) slow (3) unchanged (4) stops → answer 2)

Derivation : T = 2π √(L ÷ g)

Step 1 : The bob is displaced by θ; the tangential force F = −m g sin θ.
Step 2 : Small θ : sin θ ≈ θ = x ÷ L ⇒ F = −(m g ÷ L) x.
Step 3 : This has the form F = −k x ⇒ k = m g ÷ L.
Step 4 : T = 2π √(m ÷ k) = 2π √(L ÷ g).

Remember

T = 2π√(L/g); L ∝ T²; seconds pendulum T = 2 s ⇒ L ≈ 1 m.

Common mistake / CET trap

m has no effect. sin θ ≈ θ only in radians. L = string + radius of the bob.

Shortcut

Memory clue : "Longer = slower, stronger gravity = faster".

Spot the difference

Point	Pendulum	Spring–mass
T formula	2π√(L/g)	2π√(m/k)
Effect of m	none	yes
Effect of g	yes	none

Practice questions

Easy

1L = 0.9 m, g = 10 : T?
2T of a seconds pendulum?
3m doubled ⇒ T?

Medium

4L for T = 2 s (g = 10)?
5L 1 m → 0.25 m ⇒ T?
6Lift upward with a = g ⇒ factor on T?

Challenging

7On the Moon g = g_earth ÷ 6 ⇒ factor on T?
8Time for 100 oscillations at T = 1.5 s?

Show answers (try them yourself first)

1T = 2π √0.09 = 1.885 s
22 s
3No change
4L = 10 ÷ π² = 1.013 m
5T halves
6g' = 2g ⇒ ×1/√2
7×√6 = 2.45
8150 s

Angular SHM, torsion pendulum and vibrating magnet

From daily life

The balance wheel of a wristwatch twists to and fro — an angular oscillation.

Simple definition

Angular SHM : a body twists to and fro about an axis; the restoring torque is proportional to the angle.

In the figure : A twisted wire turns the disc by θ and the torque −Cθ pulls it back.

Rule 1: Angular SHM — The restoring torque τ = −C θ ⇒ α = −(C ÷ I) θ ⇒ ω = √(C ÷ I), T = 2π √(I ÷ C). C = torsion constant (N m/rad).

Easy : I = 2 × 10⁻³ kg m², C = 0.05 : T = 2π √0.04 = 1.257 s.
Medium : I four times ⇒ T doubles.
Exam (CET) : T ∝ √I. Unit of C : N m/rad. (Options : T ∝ 1) I 2) √I 3) 1/I 4) I² → answer 2)

Rule 2: Torsion pendulum — A disc hung from a wire and twisted performs angular SHM; C depends on the wire (length and thickness); for small angles T is independent of amplitude.

Easy : Wire twice as long ⇒ C halves ⇒ T × √2.
Medium : A thicker wire ⇒ larger C ⇒ smaller T.
Exam (CET) : Wire radius doubled ⇒ C becomes? (1) 2 (2) 4 (3) 8 (4) 16 times → answer 4). Trick : C ∝ r⁴.

Rule 3: Vibrating magnet — A bar magnet in field B : τ = −m B sin θ ≈ −m B θ ⇒ T = 2π √(I ÷ m B) (m = magnetic moment). Larger B ⇒ smaller T.

Easy : I = 10⁻⁵ kg m², m = 0.5 A m², B = 2 × 10⁻³ T : T = 2π √(10⁻⁵ ÷ 10⁻³) = 0.628 s.
Medium : B four times ⇒ T halves.
Exam (CET) : A magnet has T = 2 s; B is made 4 times : T = ? (1) 4 s (2) 1 s (3) 0.5 s (4) 2 s → answer 2)

Derivation : T = 2π √(I ÷ C)

Step 1 : Equation of angular motion : I α = τ.
Step 2 : τ = −C θ ⇒ I d²θ/dt² = −C θ.
Step 3 : d²θ/dt² = −(C ÷ I) θ ⇒ ω² = C ÷ I.
Step 4 : T = 2π ÷ ω = 2π √(I ÷ C) (m ↔ I, k ↔ C).

Remember

τ = −Cθ; T = 2π√(I/C); magnet T = 2π√(I/mB).

Common mistake / CET trap

C is in N m/rad (angle in radians). Do not mix I with m. The m in the magnet formula is the magnetic moment.

Shortcut

Memory clue : "m → I, k → C" – just swap the letters of the linear formula.

Spot the difference

Linear SHM	Angular SHM
F = −k x	τ = −C θ
m	I
T = 2π√(m/k)	T = 2π√(I/C)

Practice questions

Easy

1T of a torsion pendulum?
2Unit of C?
3Formula for the T of a vibrating magnet?

Medium

4I = 0.02, C = 0.5 : T?
5I ×9 ⇒ T?
6B ×9 ⇒ T?

Challenging

7Wire four times as long ⇒ factor on T?
8Magnetic moment ×4 ⇒ T?

Show answers (try them yourself first)

12π√(I/C)
2N m/rad
32π√(I ÷ mB)
4T = 2π √0.04 = 1.257 s
5T triples
6T ÷ 3
7C ÷ 4 ⇒ T × 2
8T halves

Damped and forced oscillations, resonance

From daily life

A swing climbs higher when pushed in the right rhythm, not when pushed at the wrong one.

Simple definition

In damped oscillations the amplitude decays; in forced oscillations an external force supplies energy; resonance is the maximum amplitude.

In the figure : The peak lies at ω_d = ω₀; light damping gives a sharp peak.

Live figure

Damped and forced oscillations, resonance

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Rule 1: Damped oscillation — The resistive force −b v makes the amplitude decay : A(t) = A₀ e^(−b t ÷ 2m); ω' = √(ω₀² − (b ÷ 2m)²). Energy is lost (as heat).

Easy : A₀ = 10 cm, b ÷ 2m = 0.1 s⁻¹, t = 10 s : A = 10 e⁻¹ = 3.68 cm.
Medium : ω₀ = 20, b ÷ 2m = 5 : ω' = √(400 − 25) = 19.36 rad/s.
Exam (CET) : In damping ω' compared to ω₀ is? (1) more (2) less (3) equal (4) zero → answer 2)

Rule 2: Forced oscillation — When an external periodic force of angular frequency ω_d acts, the system oscillates at ω_d (not at its own ω₀). The amplitude grows as ω_d nears ω₀.

Easy : Pushing a swing every 2 s : forcing frequency 0.5 Hz.
Medium : Transient oscillations first, then a steady state at ω_d.
Exam (CET) : Frequency in the steady state? (1) ω₀ (2) ω_d (3) ω' (4) 0 → answer 2)

Rule 3: Resonance — At ω_d = ω₀ the amplitude is maximum : resonance. Less damping ⇒ a sharper, taller peak. Uses : radio tuning, musical instruments; dangers : bridges, buildings.

Easy : ω₀ = √(k ÷ m) = √(40 ÷ 0.1) = 20 : resonance at ω_d = 20.
Medium : Soldiers break step on a bridge.
Exam (CET) : ω_d = 15 and ω₀ = 20 : no resonance; smaller amplitude. (Options : 1) resonance 2) none 3) infinite 4) zero energy → answer 2)

Derivation : damped ω'

Step 1 : Equation : m d²x/dt² = −k x − b dx/dt.
Step 2 : Solution x = A₀ e^(−γt) cos(ω't), with γ = b ÷ 2m.
Step 3 : Substituting gives ω'² = k ÷ m − (b ÷ 2m)².
Step 4 : ω' = √(ω₀² − (b ÷ 2m)²); larger b ⇒ smaller ω' and faster amplitude decay.

Remember

A = A₀e^(−bt/2m); ω' = √(ω₀² − (b/2m)²); resonance at ω_d = ω₀.

Common mistake / CET trap

A forced oscillation runs at ω_d, not ω₀. At resonance the amplitude is not infinite (damping limits it).

Shortcut

Memory clue : "push rhythm = natural rhythm ⇒ resonance".

Spot the difference

Type	Frequency	Amplitude
Free	ω₀	constant
Damped	ω' < ω₀	decays
Forced	ω_d	max near ω_d ≈ ω₀

Practice questions

Easy

1When does resonance occur?
2Where does energy go in damping?
3At which frequency does a forced oscillation run?

Medium

4ω₀ = 10, b ÷ 2m = 6 : ω'?
5A₀ = 8 cm, b ÷ 2m = 0.5 s⁻¹, t = 2 s : A?
6k = 90, m = 0.1 : resonance ω_d?

Challenging

7How does heavier damping change the resonance peak?
8Why is resonance dangerous on a bridge?

Show answers (try them yourself first)

1ω_d = ω₀
2As heat
3ω_d
4ω' = √(100 − 36) = 8 rad/s
5A = 8 e⁻¹ = 2.94 cm
6ω_d = √900 = 30 rad/s
7The peak is lower and broader
8The amplitude grows so much that the bridge may fail

Quick revision of the whole lesson

Concept	Key formula / fact	Unit / dimensions
SHM	a = −ω² x; x = A sin(ωt + φ); T = 2π/ω	ω : rad/s; [T⁻¹]
Velocity / acceleration	v = ω√(A² − x²); v_max = Aω; a_max = Aω²	m/s; m/s²
Energy	E = ½mω²A² = ½kA²; KE = PE ⇒ x = A/√2	J; [M L² T⁻²]
Spring	T = 2π√(m/k); k_parallel = k₁ + k₂	s; N/m
Simple pendulum	T = 2π√(L/g); seconds pendulum L ≈ 1 m	s
Angular SHM	τ = −Cθ; T = 2π√(I/C); magnet 2π√(I/mB)	N m/rad
Damping / resonance	A = A₀e^(−bt/2m); resonance at ω_d = ω₀	rad/s

🎯 Most important points for the exam

SHM : F ∝ −x, a = −ω²x, T = 2π/ω; v = ω√(A² − x²).
E = ½mω²A² is constant; KE = PE at x = A/√2.
Spring T = 2π√(m/k); pendulum T = 2π√(L/g) (no m).
Series/parallel springs; half a spring ⇒ k × 2.
In a lift g' = g ± a; no oscillation in free fall.
Angular SHM : T = 2π√(I/C).
Resonance at ω_d = ω₀; damping gives A₀e^(−bt/2m).

Study mantra

Write a = −ω²x → find ω → T = 2π/ω → pick the formula → check units.