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Thursday, October 18, 2018

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

4 comments:

  1. Thank you for the solution.

    And it's question 4 not question 7

    ReplyDelete
    Replies
    1. no problem. it's question 4 from the paper and the 7th question in the chapter Superposition present on the site.

      I am also classifying the questions topic-wise so that, with time, you may be able to access the questions after completing each chapter.
      this is available under the heading 'A-Level 9702 Topic by Topic' in the right column.

      Delete
  2. Sorry another stupid question, they said a much louder sound is first heard when X=4.5cm. so why at X=4.5cm there is a node? Shouldn't it be antinode?

    ReplyDelete
    Replies
    1. the sound is heard at the open end of the pipe. So, an antinode is formed there while a node is always formed at the piston.

      the distance is still x = 4.5 cm.

      Delete

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