wave_mechanics Beats Question 45
Question: Two strings X and Y of a sitar produce a beat frequency 4 Hz. When the tension of the string Y is slightly increased the beat frequency is found to be 2 Hz. If the frequency of X is 300 Hz, then the original frequency of Y was
[UPSEAT 2000]
Options:
A) 296 Hz
B) 298 Hz
C) 302 Hz
D) 304 Hz
Show Answer
Answer:
Correct Answer: A
Solution:
$ n _{x}=300Hz, $
$ n _{y}=? $ x = beat frequency = 4 Hz, which is decreasing (4®2) after increasing the tension of the string y.
Also tension of wire y increasing so $ n _{y}\uparrow $
$ (\because n\propto \sqrt{T}) $ Hence $ n _{x}-n _{y}\downarrow =x\downarrow $ Correct $ n _{y}\uparrow -n _{x}=x\downarrow $ Wrong
Therefore $ n _{y}=n _{x}-x=300-4=296Hz $
BETA
