Question
Two loudspeakers A and B are initially equidistant from a microphone M. The frequency and
intensity emitted by A and B are the same. A and B emit sound in phase. A is fixed in position.
B is moved slowly away from M along the line MP. The graph shows the variation with
distance travelled by B of the received intensity at M.
(a) Explain why the received intensity varies between maximum and minimum values. [3]
(b) State and explain the wavelength of the sound measured at M. [2]
(c) B is placed at the first minimum. The frequency is then changed until the received
intensity is again at a maximum.
Show that the lowest frequency at which the intensity maximum can occur is about 3kHz.
Speed of sound = 340ms-1 [2]
Answer/Explanation
Ans.
(a) movement of $B$ means that path distance is different « between $B M$ and $A M$ » OR
movement of B creates a path difference «between $B M$ and $A M » \checkmark$
interference
OR
superposition «of waves» $\checkmark$
maximum when waves arrive in phase $/$ path difference $=\mathrm{n} \times$ lambda
OR
minimum when waves arrive « $180^{\circ}$ or $\pi$ ” out of phase / path difference $=(n+1 / 2)$ x lambda $\checkmark$
(b) wavelength $=26 \mathrm{~cm} \checkmark$
peak to peak distance is the path difference which is one wavelength
$O R$
this is the distance B moves to be back in phase “with $A » \checkmark$
(c)
$
\ll \frac{\lambda}{2} n=13 \mathrm{~cm}
$
$
f=« \frac{c}{\lambda}=\frac{340}{0.13}=» 2.6 « \mathrm{kHz} » \checkmark
$