wave_mechanics critical_thinking Question 38
Question: Two loudspeakers L1 and L2 driven by a common oscillator and amplifier, are arranged as shown. The frequency of the oscillator is gradually increased from zero and the detector at D records a series of maxima and minima. If the speed of sound is 330 ms?1 then the frequency at which the first maximum is observed is
Options:
A) 165 Hz
B) 330 Hz
C) 496 Hz
D) 660 Hz
Show Answer
Answer:
Correct Answer: B
Solution:
Path difference between the wave reaching at D $ \Delta x=L _{2}P-L _{1}P $
$ =\sqrt{40^{2}+9^{2}}-40 $
$ =41-40 $ = 1m For maximum $ \Delta x=(2n)\frac{\lambda }{2} $ For first maximum (n = 1)
Therefore $ 1=2(1)\frac{\lambda }{2} $
Therefore $ \lambda =1m $
Therefore $ n=\frac{v}{\lambda }=330Hz $ .
BETA
