Class 11 Physics Chapter 14: Waves worksheet with answer key

Answer: (c) y = 0.030 sin(4πx − 80πt) m

k = 2π/λ = 2π/0.50 = 4π m⁻¹. ν = v/λ = 20/0.50 = 40 Hz, so ω = 2πν = 80π s⁻¹. A wave along +x has the form a sin(kx − ωt). The + sign gives a wave along −x; 40πt drops the 2 in ω = 2πν; 0.50x − 20t puts λ and v where k and ω belong.

Answer: (b) longitudinal, since air cannot sustain a shearing stress

Transverse waves need a medium that can sustain shearing stress, which a fluid such as air cannot. Fluids sustain compressive strain, so only longitudinal mechanical waves travel through air.

Answer: (b) 36 m s⁻¹

A phase difference of 2π corresponds to one wavelength, so π/3 corresponds to λ/6. λ = 6 × 0.15 = 0.90 m and v = λν = 0.90 × 40 = 36 m s⁻¹. 18 m s⁻¹ takes π/3 as a third of a wavelength; 6.0 m s⁻¹ multiplies the separation by the frequency.

Answer: (c) energy, while each layer of air only oscillates about its mean position

In a wave, energy and not matter is transferred. Sound travels as compressions and rarefactions; each layer of air oscillates about its mean position, unlike a wind, which is a motion of air as a whole.

Answer: (a) 15 cm

Successive nodes are λ/2 apart and an antinode lies midway between two nodes, so a node and the adjacent antinode are λ/4 = 60/4 = 15 cm apart. 30 cm is the node-to-node distance.

ν = (1/2L)√(T/μ), so ν ∝ √T and T ∝ ν². T₂/T₁ = (1.10)² = 1.21. The tension must be increased by 21%.

• ν ∝ √T, so T ∝ ν² (1)
• T₂/T₁ = 1.21: an increase of 21% (1)

The push compresses the first coupling, which pushes the second bogie; its coupling compresses and pushes the third, and so on, so the push passes from bogie to bogie along the train. Each bogie moves only a little and the train as a whole is not moved. A wave likewise carries a disturbance, with its energy, through a medium whose parts only move about their own positions; the medium as a whole does not travel.

• the push is passed on from bogie to bogie through the spring couplings (1)
• the train as a whole does not move: a wave carries the disturbance (energy), not the medium (1)

k = 2π/λ = 2π/0.40 = 5π ≈ 15.7 m⁻¹ and ω = 2πν = 2π × 25 = 50π ≈ 157 s⁻¹. A wave along −x has the form a sin(kx + ωt + φ), and φ = 0 since y = 0 at x = 0, t = 0. So y = 0.020 sin(5πx + 50πt) m.

• k = 5π m⁻¹ and ω = 50π s⁻¹ (1)
• y = 0.020 sin(5πx + 50πt), with the + sign for −x (1)

Successive resonances differ by 100 Hz. If the pipe were open at both ends, all harmonics would appear and the fundamental would be 100 Hz; but 150 Hz is not a multiple of 100 Hz. For a pipe closed at one end only odd harmonics appear, spaced by 2ν₁: ν₁ = 100/2 = 50 Hz, and 150, 250, 350 Hz are its 3rd, 5th and 7th harmonics. So the pipe is closed at one end, and L = v/4ν₁ = 340/(4 × 50) = 1.7 m.

• closed at one end: the frequencies are in the ratio 3 : 5 : 7 (1)
• fundamental 50 Hz (1)
• L = v/4ν₁ = 1.7 m (1)

A point of fixed phase, such as a crest, moves so that kx − ωt stays constant. If it moves Δx in time Δt, k(x + Δx) − ω(t + Δt) = kx − ωt, so kΔx − ωΔt = 0 and, in the limit, dx/dt = ω/k = v. With ω = 2πν and k = 2π/λ, v = 2πν/(2π/λ) = λν = λ/T. Physically: in the time an element of the medium takes for one full oscillation, the wave pattern travels one wavelength.

• kx − ωt = constant for a moving phase point (1)
• kΔx = ωΔt, so v = ω/k (1)
• ω = 2πν, k = 2π/λ give v = λν; meaning: the pattern moves λ in one period (1)

(a) At a rigid end the displacement must be zero at all times; by superposition this needs a reflected pulse differing in phase by π, so the pulse returns inverted, of the same shape and height if no energy is absorbed (the wall pulls back on the string, by Newton's third law). (b) At a free end there is no phase change: the pulse returns upright, with the same height. The incident and reflected pulses add at the end, so the end rises to twice the pulse height, 10 mm.

• rigid end: inverted, phase change π, with reason (1)
• free end: upright, no phase change (1)
• greatest displacement of the free end 2 × 5 = 10 mm (1)