Wave Motion Exercise (With comprehensive Solutions)

Wave Motion Exercise Solutions

Question 1

(a)(i) A string has mass per unit length of 0.05 kg/m, calculate the tension in the string along which vibrations have a speed of 8 cm/s.

We use the wave speed formula:

v = √(T/μ)

Where:
v = 0.08 m/s (wave speed)
μ = 0.05 kg/m (linear mass density)

Rearranged: T = v² × μ = (0.08)² × 0.05 = 0.00032 N

Answer: 3.2 × 10⁻⁴ N

(a)(ii) Two forks, A and B, when sounded together produce 4 beats/second. The fork A is in unison with 30 cm length of a sonometer wire and B is in unison with 25 cm length of the same wire at the same tension. Calculate the frequencies of the forks.

For a sonometer wire, frequency is inversely proportional to length:

f ∝ 1/L

Let f_A = frequency of fork A, f_B = frequency of fork B

f_A/f_B = L_B/L_A = 25/30 = 5/6

Beat frequency = |f_A - f_B| = 4 Hz

Let f_A = 5x, f_B = 6x ⇒ 6x - 5x = 4 ⇒ x = 4

Therefore: f_A = 20 Hz, f_B = 24 Hz

Answer: Fork A = 20 Hz, Fork B = 24 Hz

(b) A wave travelling along a string is described by y(x,t) = 3.35 sin(2.7t - 72.1x) where y is in mm, x in meters, t in seconds. Find:

(i) Amplitude

Amplitude is the coefficient of the sine function: 3.35 mm

Answer: 3.35 mm

(ii) Wavelength

k = 72.1 m⁻¹ (wave number)
λ = 2π/k = 2π/72.1 ≈ 0.087 m

Answer: 0.087 m

(iii) Period

ω = 2.7 rad/s (angular frequency)
T = 2π/ω = 2π/2.7 ≈ 2.33 s

Answer: 2.33 s

(iv) Frequency

f = 1/T = 1/2.33 ≈ 0.43 Hz

Answer: 0.43 Hz

(v) Velocity

v = ω/k = 2.7/72.1 ≈ 0.0374 m/s = 3.74 cm/s

Answer: 3.74 cm/s

(c) Two identical waves travelling in opposite directions form: y = 8 cos(2x) sin(3t). Find the particle displacements.

Using the standing wave identity:

2A cos(kx) sin(ωt) = A sin(ωt - kx) + A sin(ωt + kx)

Comparing with given equation (A = 4, k = 2, ω = 3):

y₁ = 4 sin(3t - 2x)

y₂ = 4 sin(3t + 2x)

Answer: y₁ = 4 sin(3t - 2x), y₂ = 4 sin(3t + 2x)

(d) Standing wave: y = 0.35 sin(0.25x) cos(12πt). Find wavelength, frequency, amplitude.

From the standing wave equation:

Amplitude of progressive waves: A = 0.35/2 = 0.175 m

Wave number k = 0.25 m⁻¹ ⇒ λ = 2π/k = 2π/0.25 ≈ 25.13 m

Angular frequency ω = 12π rad/s ⇒ f = ω/2π = 6 Hz

Answer: λ = 25.13 m, f = 6 Hz, A = 0.175 m

Question 2

(a) Wire: length 75 cm, mass 16.5 g. Find tension needed for λ=3.33 cm waves at 875 Hz.

First find linear mass density:

μ = mass/length = 0.0165 kg/0.75 m = 0.022 kg/m

Wave speed v = fλ = 875 × 0.0333 ≈ 29.14 m/s

Tension T = v²μ = (29.14)² × 0.022 ≈ 18.68 N

Answer: 18.7 N

(b) Piano wire: tension 800 N, length 0.40 m, mass 3.0 g.

(i) Fundamental frequency

μ = 0.003/0.4 = 0.0075 kg/m
f₁ = (1/2L)√(T/μ) = (1/0.8)√(800/0.0075) ≈ 408.25 Hz

Answer: 408.25 Hz

(ii) Highest audible harmonic (up to 10207 Hz)

n = 10207/408.25 ≈ 25

Answer: 25th harmonic

(c) String: length 2 m, mass 6.0 × 10⁻⁴ kg, tension 20 N, plucked 20 cm from end. Find frequency.

μ = 6.0 × 10⁻⁴/2 = 3.0 × 10⁻⁴ kg/m
f₁ = (1/2L)√(T/μ) = (1/4)√(20/3.0 × 10⁻⁴) ≈ 129.1 Hz

But plucking position affects which harmonics are excited. The fundamental dominates:

Answer: 129.1 Hz (fundamental frequency)

(d) String produces 10 beats/s at 129.6 N tension, matches fork at 160 N. Find fork frequency.

Frequency ∝ √Tension

At 160 N: f = f₀ (matches fork)

At 129.6 N: f' = f₀√(129.6/160) = 0.9f₀

Beat frequency: f₀ - f' = 0.1f₀ = 10 ⇒ f₀ = 100 Hz

Answer: 100 Hz

Question 3

(a) Derive relationship between sound velocity V, Young's modulus Y, and density ρ.

Using dimensional analysis:

Assume V = kYᵃρᵇ

Dimensions: [V]=[LT⁻¹], [Y]=[ML⁻¹T⁻²], [ρ]=[ML⁻³]

Solving: a = ½, b = -½

V = k√(Y/ρ)

Answer: V = √(Y/ρ)

(b) Calculate sound velocity in:

(i) Steel: ρ=7800 kg/m³, Y=2.0×10¹¹ N/m²

V = √(Y/ρ) = √(2.0×10¹¹/7800) ≈ 5064 m/s

Answer: 5064 m/s

(ii) Water: ρ=1000 kg/m³, B=2.04×10⁹ N/m²

V = √(B/ρ) = √(2.04×10⁹/1000) ≈ 1428 m/s

Answer: 1428 m/s

Question 4

(a) Calculate sound velocity in air at 100°C (STP density=1.29 kg/m³, γ=1.41).

V = √(γP/ρ)
At STP (0°C): V₀ = 331 m/s
At 100°C: V = V₀√(373/273) ≈ 398 m/s

Answer: 398 m/s

(b) For λ=66.5 cm at 17°C (512 Hz), find γ of air.

V = fλ = 512 × 0.665 ≈ 340.5 m/s

Using V = √(γP/ρ) and knowing P/ρ at STP:

γ ≈ (V²ρ)/P ≈ 1.39

Answer: 1.39

(c) Draw diagram and calculate frequency for 2nd overtone in 72 cm closed pipe (V=340 m/s).

Diagram: Closed pipe has node at closed end, antinode at open end

2nd overtone = 5th harmonic: L = 5λ/4 ⇒ λ = 4L/5 = 0.576 m

f = V/λ = 340/0.576 ≈ 590.28 Hz

Answer: 590.28 Hz

(d) Closed pipe (400 mm) resonates at 215 Hz minimum.

(i) Sound speed

Fundamental: L = λ/4 ⇒ λ = 4L = 1.6 m

V = fλ = 215 × 1.6 = 344 m/s

Answer: 344 m/s

(ii) Next resonance frequency

Next harmonic is 3rd: f₃ = 3f₁ = 3 × 215 = 645 Hz

Answer: 645 Hz

Question 5

(a) Open pipe (30 cm) and closed pipe (23 cm) in unison at 1st overtone. Find end correction.

For open pipe (1st overtone): f = 2V/(2(L₀ + e))

For closed pipe (1st overtone): f = 3V/(4(L_c + e))

Setting equal and solving: e ≈ 0.01 m = 1 cm

Answer: 1 cm

(b) Two open pipes sound harmonics together. First pipe is 0.80 cm, find second pipe length.

When harmonics coincide:

n₁/L₁ = n₂/L₂

For simplest case (fundamentals): L₂ = L₁ = 0.80 cm

Or could be other harmonic combinations (needs more info)

Answer: 0.80 cm (assuming fundamental frequencies match)