Progressive, transverse/longitudinal and stationary waves (RP1) – A-level Physics exam questions with mark scheme

1Core3 marks

A progressive wave has a frequency of 50 Hz and a wavelength of 0.60 m.

(a) Calculate the phase difference, in radians, between two points on the wave that are 0.15 m apart along the direction of travel. [2]

(b) Calculate the speed of the wave. [1]

(a) phase difference = $\dfrac{0.15}{0.60} \times 2\pi$(1)
(a) $\dfrac{\pi}{2}$ rad allow 1.57 rad(1)
(b) $c = f\lambda = 50 \times 0.60 = 30$ m s$^{-1}$(1)
Answer: (a) $\pi/2$ rad (b) 30 m s$^{-1}$
Examiner tip: Phase difference = (distance ÷ wavelength) × 2π; giving the answer in degrees when radians are asked costs the mark.

2Core3 marks

(a) State what is meant by a plane-polarised wave. [1]

(b) Explain why sound waves cannot be polarised. [1]

(a) oscillations (of the field / particles) are in one plane only (containing the direction of travel) allow vibrations; allow 'oscillations in one direction only'; reject 'wave travels in one plane / one direction'(1)
(b) sound is longitudinal: the oscillations are parallel to the direction of travel (energy transfer), so there is no perpendicular oscillation to restrict 'sound is longitudinal' alone is enough for the mark(1)
(c) e.g. aligning a TV / radio aerial with the transmitted signal; Polaroid (polarising) sunglasses reducing glare; polarising filters in photography; stress analysis of plastics allow any other valid application, e.g. LCD screens, 3D cinema glasses; reject 'sunglasses' without reference to Polaroid / polarising(1)
Answer: (a) oscillations in one plane only (b) sound is longitudinal (c) e.g. aligning TV aerials
Examiner tip: Say 'oscillations in one plane', not 'waves in one direction' — direction of travel is not what polarisation restricts.

3Standard6 marks

A string is stretched between a vibration generator and a pulley. The vibrating length of the string is 0.80 m and the mass per unit length is $1.2\times10^{-3}$ kg m$^{-1}$. A mass of 0.50 kg hangs from the end of the string.

(a) Calculate the frequency of the first harmonic. [2]

(b) The frequency is increased until the third harmonic is seen. State this frequency and the number of nodes on the string, including both ends. [2]

(a) $T = 0.50 \times 9.81 = 4.9$ N and $f = \dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}} = \dfrac{1}{1.60}\sqrt{\dfrac{4.905}{1.2\times10^{-3}}}$(1)
(a) 40 Hz(1)
(b) 120 Hz ecf 3 × their (a)(1)
(b) 4 nodes(1)
(c) waves reflect at the ends, so two waves of the same frequency (and similar amplitude) travel in opposite directions and superpose(1)
(c) nodes form where the waves always cancel (destructive interference); antinodes where they reinforce(1)
Answer: (a) 40 Hz (b) 120 Hz, 4 nodes (c) superposition of incident and reflected waves
Examiner tip: Tension is the weight of the hanging mass (mg), not the mass itself — forgetting × 9.81 gives a frequency about 3 times too small.

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