(a) For a progressive wave, state what is meant by (i) the period, (ii) the wavelength. (b) Fig. 4.1 shows the variation with time t

Q#19 (Past Exam Paper – June 2018 Paper 21 Q4)

(a) For a progressive wave, state what is meant by

(i) the period, [1]

(ii) the wavelength. [1]

(b) Fig. 4.1 shows the variation with time t of the displacement x of two progressive waves P and Q passing the same point.

Fig. 4.1

The speed of the waves is 20 cm s^-1.

(i) Calculate the wavelength of the waves. [2]

(ii) Determine the phase difference between the two waves. [1]

(iii) Calculate the ratio

intensity of wave Q

intensity of wave P

[2]

(iv) The two waves superpose as they pass the same point. Use Fig. 4.1 to determine the resultant displacement at time t = 0.45 s. [1]

[Total: 8]

Solution:

(a)

(i) The period is the time for one oscillation.

(ii) The wavelength is the minimum distance between two points in phase (having the same displacement and moving in the same direction).

(b)

(i)

v = λ / T                       or v = f λ  and f = 1 / T

{λ = v × T

From the graph, the period is 0.60 s.}

λ = 20 × 0.60

λ = 12 cm

(ii)

{Consider the initial position of wave P (at equilibrium position). This is point (0,0).

Wave Q only reaches the equilibrium position at time t = 0.20 s.

The difference between the 2 waves is 0.20 s.

1 period corresponds to a phase difference of 360°. We need to find the phase difference that corresponds to a time of 0.20 s.

0.60 s - - > 360°

0.20 - - > 360° × (0.20 / 0.60)

Alternatively, (since we do not know the direction in which the waves are moving {left or right}) we could say that the ‘trough’ of wave Q occurs before that of wave P. The difference in time is 0.40 s.

Here, we find the phase difference corresponding to 0.40 s.}

phase difference = 360° × (0.20 / 0.60)                     or 360° × (0.40 / 0.60)

phase difference = 120°                                             or 240°

(iii)

{Intensity is proportional to (amplitude)².}

I ∝ A²  

{I_P ∝ 3²

I_Q ∝ 2²}

IQ / IP = AQ² / AP²

IQ / IP = 2.0² / 3.0²

IQ / IP = 0.44

(iv)

{Displacement is a vector quantity, so we need to consider the direction.

From the graph, at t = 0.45 s

Displacement of point on wave P = – 3.00 mm

Displacement of point on wave Q = 1.00 mm}

displacement = 1.00 – 3.00

displacement = –2.00 mm

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