wave_mechanics Beats Question 7
Question: The frequencies of two sound sources are 256 Hz and 260 Hz. At t = 0, the intensity of sound is maximum. Then the phase difference at the time t = 1/16 sec will be
Options:
A) Zero
p
p/2
p/4
Show Answer
Answer:
Correct Answer: C
Solution:
Time interval between two consecutive beats $ T=\frac{1}{n _{1}-n _{2}}=\frac{1}{260-256}=\frac{1}{4}\sec $ so, $ t=\frac{1}{16}=\frac{T}{4}\sec $ By using time difference =$ \frac{T}{2\pi }\times $ Phase difference $ \Rightarrow $ $ \frac{T}{4}=\frac{T}{2\pi }\times \varphi \Rightarrow \varphi =\frac{\pi }{2} $
BETA
