State the difference between a stationary wave and a progressive wave in terms of

Question 32

(a) State the difference between a stationary wave and a progressive wave in terms of

(i) the energy transfer along the wave, [1]

(ii) the phase of two adjacent vibrating particles. [1]

(b) A tube is open at both ends. A loudspeaker, emitting sound of a single frequency, is placed near one end of the tube, as shown in Fig. 3.1.

Fig. 3.1

The speed of the sound in the tube is 340 m s^-1. The length of the tube is 0.60 m.

A stationary wave is formed with an antinode A at each end of the tube and two antinodes inside the tube.

(i) State what is meant by an antinode of the stationary wave. [1]

(ii) State the distance between a node and an adjacent antinode. [1]

(iii) Determine, for the sound in the tube,

1. the wavelength, [1]

2. the frequency. [2]

(iv) Determine the minimum frequency of the sound from the loudspeaker that produces a stationary wave in the tube. [2]

 [Total: 9]

Reference: Past Exam Paper – November 2017 Paper 21 Q3

Solution:

(a)

(i) In a stationary wave energy is not transferred while in a progressive wave energy is transferred

(ii) In a stationary wave (adjacent) particles are in phase while in a progressive wave (adjacent) particles are out of phase

(b)

(i) An antinode is (position where) maximum amplitude

(ii) distance = 0.10 m

{From the diagram,

Separation between 3 AA = 0.60 m

Separation between 1 AA = 0.60 / 3 = 0.20 m

The distance between a node and an antinode is half the distance between AA.

Distance = 0.20 / 2 = 0.10 m}

(iii)

1.

{The stationary wave formed in the tube corresponds to 1.5 wavelength.

Note that the distance between 2 antinodes is 0.5 wavelength.

1.5 λ = 0.60}

λ = 0.60 / 1.5

λ = 0.40 m

2.

v = fλ              

f = 340 / 0.40

f = 850 Hz

(iv)

{The tube is opened at both ends. So, antinodes are formed at the ends of the tube.

The fundamental stationary wave consists of only an antinode at each end of the tube. This corresponds to half a wavelength.

½ λ = 0.60 m}

λ  = 2 × 0.60

{f = v / λ = 340 / 1.2 = 283 Hz}

f = 280 (283) Hz