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
Thank you for the solution.
ReplyDeleteAnd it's question 4 not question 7
no problem. it's question 4 from the paper and the 7th question in the chapter Superposition present on the site.
DeleteI 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.
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?
ReplyDeletethe 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.
Deletethe distance is still x = 4.5 cm.