Exercises
1: Which of the following represent periodic motion? (a) Swimmer trip (b) Bar magnet released (c) H₂ molecule rotating (d) Arrow from bow.
Answer
(a) Swimmer: Not periodic — it's a one-time return trip, not repeating.
(b) Bar magnet displaced and released: Periodic motion — the magnet oscillates about its N-S equilibrium direction with a definite period.
(c) H₂ molecule rotating about its centre of mass: Periodic motion — the rotation repeats with a definite period.
(d) Arrow from bow: Not periodic — it moves in one direction and doesn't return to its original position.
2: Which represent (nearly) simple harmonic motion and which are periodic but not SHM?
Answer
(a) Rotation of Earth about axis: Periodic but not SHM — it's uniform circular motion, not oscillation about an equilibrium position.
(b) Oscillating mercury column in U-tube: SHM — restoring force is proportional to displacement from equilibrium.
(c) Ball bearing in smooth curved bowl (small oscillations): SHM — for small displacements, the restoring force is proportional to displacement.
(d) General vibrations of polyatomic molecule: Periodic but not SHM — superposition of many different vibrational modes gives complex non-sinusoidal motion.
3: Four x-t plots — which represent periodic motion? State periods.
Answer
(a) Not periodic — the displacement increases monotonically.
(b) Periodic — sinusoidal motion with period T = 2 s.
(c) Not periodic — displacement is erratic and non-repeating.
(d) Periodic — repeats with T = 2 s (even though not sinusoidal; it is periodic but not SHM).
4: Classify each function as (a) SHM, (b) periodic but not SHM, (c) non-periodic. Give period if periodic.
Answer
(a) sin ωt − cos ωt:
= √2 sin(ωt − π/4)
⇒ SHM, period T = 2π/ω
(b) sin³ ωt:
sin³ ωt = (3 sin ωt − sin 3ωt)/4
⇒ superposition of two SHMs with frequencies ω and 3ω
⇒ Periodic but not SHM, period T = 2π/ω
(c) 3 cos(π/4 − 2ωt):
= 3 cos(2ωt − π/4)
⇒ SHM, period T = 2π/(2ω) = π/ω
(d) cos ωt + cos 3ωt + cos 5ωt:
Sum of SHMs with different frequencies
⇒ Periodic but not SHM, period T = 2π/ω (LCM period)
(e) exp(−ω²t²)
Decays monotonically → Non-periodic
(f) 1 + ωt + ω²t²
Polynomial — increases monotonically → Non-periodic
5: Particle in SHM between A and B, 10 cm apart. Give signs of velocity, acceleration, force at various positions.
Answer
Take A at −5 cm, B at +5 cm, centre O at 0. Direction A
⇒ B is positive.
Position | Velocity | Acceleration | Force |
(a) At end A (x = −5 cm) | Zero | +ve (towards O) | +ve |
(b) At end B (x = +5 cm) | Zero | −ve (towards O) | −ve |
(c) Mid-point, going towards A | −ve | Zero (a = −ω²x = 0 at x=0) | Zero |
(d) 2 cm from B towards A (x = +3 cm) | −ve | −ve | −ve |
(e) 3 cm from A towards B (x = −2 cm) | +ve | +ve | +ve |
(f) 4 cm from B towards A (x = +1 cm) | −ve | −ve | −ve |
6: Which relationships between acceleration a and displacement x involve SHM?
Answer
SHM requires: a = −ω²x (restoring force proportional to x, directed towards equilibrium, i.e., negative sign)
(a) a = 0.7x: Proportional to x but positive sign
⇒ not SHM (unstable equilibrium)
(b) a = −200x²: Not proportional to x (depends on x²)
⇒ not SHM
(c) a = −10x: a ∝ −x → SHM with ω² = 10
(d) a = 100x³: Depends on x³ → not SHM
7: x(t) = A cos(ωt + φ). At t = 0: x = 1 cm, v = ω cm/s, ω = π s⁻¹. Find A and φ.
Answer
At t = 0: x(0) = A cos φ = 1 cm ... (1)
v(0) = −Aω sin φ = ω ...
⇒ −A sin φ = 1
⇒ A sin φ = −1 ... (2)
From (1)² + (2)²: A² = 1 + 1 = 2
⇒ A = √2 cm
From (1): cos φ = 1/√2
⇒ φ = π/4 or −π/4
From (2): sin φ = −1/√2
⇒ φ = −π/4 (fourth quadrant)
φ = −π/4 radians
For sine form x = B sin(ωt + α)
B sin α = 1 and B cos α = 1 → B = √2 cm, α = π/4
8: Spring balance reads 0 to 50 kg, scale length 20 cm. Body oscillates with T = 0.6 s. Find weight of body.
Answer
Spring constant: Maximum force = 50 × 9.8 = 490 N for extension 0.20 m
k = F/x = 490/0.20 = 2450 N/m
Period of oscillation: T = 2π√(m/k)
⇒ m = k(T/2π)²
m = 2450 × (0.6/2π)² = 2450 × (0.09549)² = 2450 × 9.12 × 10⁻³ = 22.34 kg
Weight = mg = 22.34 × 9.8 = 219 N ≈ 219 N
9: Spring (k = 1200 N/m), mass 3 kg, pulled 2.0 cm and released. Find (i) frequency, (ii) max acceleration, (iii) max speed.
Answer
(i) Frequency: ω = √(k/m) = √(1200/3) = √400 = 20 rad/s
f = ω/(2π) = 20/(2π) = 3.18 Hz ≈ 3.2 Hz
(ii) Maximum acceleration: amax = ω²A = 400 × 0.02 = 8 m/s² (at extreme position)
(iii) Maximum speed: vmax = ωA = 20 × 0.02 = 0.4 m/s (at equilibrium)
10: Same spring-mass system. Write x(t) for three initial conditions.
Answer
ω = 20 rad/s, A = 2.0 cm = 0.02 m
(a) At mean position (x = 0) at t = 0, moving in +x direction:
x(t) = A sin(ωt) = 0.02 sin(20t) m
(b) At maximum stretched position (x = +A) at t = 0:
x(t) = A cos(ωt) = 0.02 cos(20t) m
(c) At maximum compressed position (x = −A) at t = 0:
x(t) = −A cos(ωt) = −0.02 cos(20t) m
All three have same frequency and amplitude, but different initial phases (0, 0, π respectively).
11: Circular motion reference — obtain SHM for x-projection of radius vector.
Answer
For circular motion, the projection of the radius vector on any axis executes SHM.
If a particle moves anticlockwise with angular speed ω and radius r, starting at angle φ₀:
x(t) = r cos(ωt + φ₀)
The specific form depends on the values in Fig. 13.20 (radius, period, initial position, direction):
Case (a): x(t) = 3 sin(πt) cm [anticlockwise, starting on y-axis, r=3 cm, T=2s]
Case (b): x(t) = 2 cos(2πt/T + π) cm [as shown in the figure]
12: Plot reference circles for the given SHM functions.
Answer
(a) x = −2 sin(3t + π/3) = 2 cos(3t + π/3 + π/2) = 2 cos(3t + 5π/6)
Radius = 2 cm, ω = 3 rad/s, initial angle = 5π/6 (150°) from x-axis, anticlockwise.
(b) x = cos(π/6 − t) = cos(t − π/6)
Radius = 1 cm, ω = 1 rad/s, initial angle = −π/6 (−30°) from x-axis.
(c) x = 3 sin(2πt + π/4) = 3 cos(2πt + π/4 − π/2) = 3 cos(2πt − π/4)
Radius = 3 cm, ω = 2π rad/s, initial angle = −π/4 (−45°).
(d) x = 2 cos(πt)
Radius = 2 cm, ω = π rad/s, initial angle = 0°.
13: Spring clamped at one end with mass m; same spring with masses at both ends. Maximum extension and period in each case.
Answer
(a) Maximum extension:
Fig (a): Force F on mass m
⇒ extension x = F/k in each case
Fig (b): Both sides stretch by F/k each. Total extension = 2F/k... but wait — each half of the spring has effective spring constant 2k, so each extends F/(2k), total = F/k. Actually, when both ends are pulled by F: total extension = 2F/k (since the spring length doubles the force).
Fig (a): xmax = F/k; Fig (b): xmax = 2F/k
(b) Period of oscillation:
Fig (a): T = 2π√(m/k)
Fig (b): The system's centre of mass is fixed; each mass oscillates as if on a spring of effective constant 2k (reduced mass system). T = 2π√(m/2k) = (1/√2) × 2π√(m/k)
Alternatively: T = 2π√(μ/k) where μ = m·m/(m+m) = m/2
⇒ T = 2π√(m/2k)
14: Piston SHM, stroke 1.0 m, ω = 200 rad/min. Maximum speed?
Answer
Stroke = 2A = 1.0 m → A = 0.5 m
ω = 200 rad/min = 200/60 rad/s = 10/3 rad/s
vmax = ωA = (10/3) × 0.5 = 5/3 = 1.67 m/s
15: Simple pendulum: T = 3.5 s on Earth. Period on Moon? gmoon = 1.7 m/s², gearth = 9.8 m/s².
Answer
T = 2π√(L/g) → T ∝ 1/√g
Tmoon/Tearth = √(gearth/gmoon) = √(9.8/1.7) = √5.765 = 2.401
Tmoon = 3.5 × 2.401 = 8.40 s
16: Pendulum (mass M, length l) in car moving at uniform speed v on circular track radius R. Time period?
Answer
The car provides centripetal acceleration ac = v²/R directed radially inward (horizontal).
Effective gravitational acceleration: geff = √(g² + ac²) = √(g² + v⁴/R²)
T = 2π√(l/geff) = 2π√[l / √(g² + v⁴/R²)]
17: Cork floats (density ρ, base area A, height h) in liquid (density ρl). Depressed and released. Show T = 2π√(hρ/ρlg).
Answer
At equilibrium, buoyancy = weight: ρlAx₀g = ρAhg
⇒ x₀ = ρh/ρl
If depressed by x: Net restoring force = −ρlAgx (excess buoyancy)
By Newton's second law: ρAh · ẍ = −ρlAgx
ẍ = −(ρlg/ρh)x → ω² = ρlg/(ρh)
T = 2π/ω = 2π√(ρh/ρlg) ✓
18: Mercury in U-tube, pressure difference maintained, pump removed. Show column executes SHM.
Answer
Let x = displacement of mercury from equilibrium (one column rises by x, other falls by x).
Net restoring force = weight of mercury column of height 2x = ρ·A·2x·g (directed opposite to displacement)
By Newton's second law: M·ẍ = −2ρAgx, where M = ρAL (total mercury mass, L = total length)
ρAL·ẍ = −2ρAgx → ẍ = −(2g/L)x
This is of the form ẍ = −ω²x with ω² = 2g/L
⇒ SHM with T = 2π√(L/2g) ✓
Important Extra Questions
Extra Q1: Define SHM. Write the equation of SHM and identify the physical meaning of each term.
Answer:
Simple Harmonic Motion (SHM) is the oscillatory motion of a particle about a fixed equilibrium point, where the restoring force is directly proportional to the displacement from equilibrium and directed towards it.
Equation: x(t) = A cos(ωt + φ)
- A = amplitude (maximum displacement from equilibrium, in m)
- ω = angular frequency (= 2πf = 2π/T, in rad/s)
- t = time (in s)
- φ = initial phase (phase constant, depends on initial conditions)
- (ωt + φ) = phase at time t
Velocity: v = −Aω sin(ωt + φ), max = Aω
Acceleration: a = −Aω² cos(ωt + φ) = −ω²x, max = Aω²
Extra Q2: Derive the expression for the time period of a simple pendulum. State the conditions under which it is valid.
Answer:
For a pendulum (bob mass m, string length L), the restoring force is the tangential component of gravity:
F = −mg sin θ ≈ −mg θ = −mg(x/L) for small angles
By Newton's law: mẍ = −(mg/L)x → ẍ = −(g/L)x
Comparing with ẍ = −ω²x: ω² = g/L → T = 2π√(L/g)
Conditions for validity
- Small amplitude (θ < 15°) so that sin θ ≈ θ.
- String is massless and inextensible.
- Bob is a point mass.
- No air resistance or friction.
Extra Q3: Derive expressions for kinetic energy and potential energy in SHM. Show total energy is constant.
Answer:
For SHM: x = A cos(ωt + φ), v = −Aω sin(ωt + φ)
Kinetic Energy: K = (1/2)mv² = (1/2)mA²ω² sin²(ωt + φ) = (1/2)kA² sin²(ωt + φ)
Potential Energy: U = (1/2)kx² = (1/2)kA² cos²(ωt + φ)
Total Energy: E = K + U = (1/2)kA²[sin²(ωt + φ) + cos²(ωt + φ)] = (1/2)kA² = constant
Total energy depends only on amplitude — it does not change with time. Energy oscillates between KE and PE, but their sum remains constant.
Extra Q4: What is resonance? Give two practical examples — one beneficial, one harmful.
Answer:
Resonance occurs when the frequency of an external periodic driving force matches the natural frequency of a system. At resonance, the amplitude of oscillation becomes very large (theoretically infinite without damping).
Condition: fdriving = fnatural
Beneficial example — Musical instruments: The body of a guitar resonates at the frequencies produced by plucked strings, amplifying the sound. Radio tuning circuits resonate at specific frequencies to select radio stations.
Harmful example — Tacoma Narrows Bridge collapse (1940): Wind-induced periodic forces matched the natural frequency of the bridge, causing resonant oscillations that grew until the bridge collapsed. Engineers now carefully design bridges to avoid resonance with wind frequencies.
Extra Q5: A 5 kg block attached to a spring (k = 500 N/m) oscillates on a frictionless surface. At t = 0, x = 0.1 m and v = 0. Find: (a) period, (b) maximum speed, (c) speed at x = 0.05 m.
Answer:
Given: m = 5 kg, k = 500 N/m, A = 0.1 m (since v = 0 at x = A)
(a) Period: T = 2π√(m/k) = 2π√(5/500) = 2π√(0.01) = 2π × 0.1 = 0.628 s ≈ 0.63 s
(b) Maximum speed: vmax = ωA = √(k/m) × A = √(500/5) × 0.1 = 10 × 0.1 = 1.0 m/s
(c) Speed at x = 0.05 m: Using energy conservation:
(1/2)mv² = (1/2)kA² − (1/2)kx²
v² = (k/m)(A² − x²) = 100 × (0.01 − 0.0025) = 100 × 0.0075 = 0.75
v = √0.75 = 0.866 m/s ≈ 0.87 m/s