# CIE AS & A Level Physics : 7.4 Electromagnetic spectrum – Exam style question – Paper 2

### Question

A long rope is held under tension between two points A and B. Point A is made to vibrate vertically and a wave is sent down the rope towards B as shown in Fig. 5.1. The time for one oscillation of point A on the rope is 0.20 s. The point A moves a distance of 80 mm during one oscillation. The wave on the rope has a wavelength of 1.5 m.
(a) (i) Explain the term displacement for the wave on the rope.

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(ii) Calculate, for the wave on the rope,
1. the amplitude,
amplitude = …………………………………… mm 

2. the speed.

speed = …………………………………. m/ s

(b) On Fig. 5.1, draw the wave pattern on the rope at a time 0.050 s later than that shown.

(c) State and explain whether the waves on the rope are

(i) progressive or stationary,

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(ii) longitudinal or transverse.

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Ans:

(a)(i) displacement is the distance the rope / particles are (above or below) from the equilibrium / mean / rest / undisturbed position (not ‘distance moved’)

(ii) 1. amplitude (= 80 / 4) = 20 mm
2. v = fλ or v = λ / T
f = 1 / T = 1 / 0.2 (5 Hz)
v = 5 × 1.5 = 7.5 m/s
(b) point A of rope shown at equilibrium position
same wavelength, shape, peaks / wave moved 1⁄4λ to right

(c)(i) progressive as energy OR peaks OR troughs is/are transferred/moved /propagated (by the waves)

(ii) transverse as particles/rope movement is perpendicular to direction of travel /propagation of the energy/wave velocity

### Question

A source of radio waves sends a pulse towards a reflector. The pulse returns from the reflector and is detected at the same point as the source. The emitted and reflected pulses are recorded on a cathode-ray oscilloscope (c.r.o.) as shown in Fig. 2.1. The time-base setting is $$0.20 μs cm^{–1}$$.
(a) Using Fig. 2.1, determine the distance between the source and the reflector.

distance = ……………………………………… m 

(b) Determine the time-base setting required to produce the same separation of pulses on the c.r.o. when sound waves are used instead of radio waves.
The speed of sound is 300 m/ s.

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Ans:

(a) $$d = v × t$$
$$t = 0.2 × 4$$ (allow $$t = 0.2 × 2$$)
$$d = 3 × 108 × 0.8 × 10^{–6}$$ OR $$3 × 108 × 0.4 × 10^{–6}$$
d = 240 m hence distance from source to reflector = 120 m

(b) speed of sound 300 cf speed of light $$3 × 10^8$$             OR               time $$= 240 / 300 (= 0.8)$$
OR              time $$= 120 / 300 (= 0.4)$$

sound slower by factor of $$10^6$$                                               OR              time for one division $$0.8 / 4$$
OR                 time for one division $$0.4 / 2$$
time base setting 0.2 s cm–1 [unit required]

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