A loudspeaker producing sound of constant frequency is placed near the open end of a pipe, as shown in Fig. 4.1.

Question 7

(a) (i) Define the wavelength of a progressive wave. [1]

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

(b) A loudspeaker producing sound of constant frequency is placed near the open end of a pipe, as shown in Fig. 4.1.

Fig. 4.1

A movable piston is at distance x from the open end of the pipe. Distance x is increased from x = 0 by moving the piston to the left with a constant speed of 0.75 cm s^-1.

The speed of the sound in the pipe is 340 m s^-1.

(i) A much louder sound is first heard when x = 4.5 cm. Assume that there is an antinode of a stationary wave at the open end of the pipe.

Determine the frequency of the sound in the pipe. [3]

(ii) After a time interval, a second much louder sound is heard. Calculate the time interval between the first louder sound and the second louder sound being heard. [2]

[Total: 7]

Reference: Past Exam Paper – June 2018 Paper 22 Q4

Solution:

(a)

(i) The wavelength is the distance between two adjacent (consecutive) wavefronts.

(ii) An antinode is a position where the stationary wave has maximum amplitude.

(b)

(i)

{There is an antinode of the stationary wave at the open end and a node at the surface of the piston.

So, x is the distance between an adjacent node and antinode. This is equal to a quarter of a wavelength.

λ / 4 = 0.45 m}

λ = 4 × 0.045

λ = 0.18 (m) or 18 (cm))

v = fλ                          

{We consider the speed of the wave in this equation, not the speed of the piston.}

f = 340 / 0.18

f = 1900 Hz    

(ii)

{The distance between 2 adjacent antinodes is half a wavelength.}

distance = λ / 2

distance = 0.09 (m) or 9 (cm))

{Speed = distance / time}

time = 0.09 / 0.0075 = 12 s