Wave Mechanics Question 755
Question: A tuning fork of frequency 392 Hz, resonates with 50 cm length of a string under tension (T). If length of the string is decreased by 2%, keeping the tension constant, the number of beats heard when the string and the tuning fork made to vibrate simultaneously is
[BHU 2004]
Options:
A) 4
B) 6
C) 8
D) 12
Show Answer
Answer:
Correct Answer: C
Solution:
$ n\propto \frac{1}{l} $
Therefore $ \frac{\Delta n}{n}=-\frac{\Delta l}{l} $ If length is decreased by 2% then frequency increases by 2% i.e., $ \frac{n _{2}-n _{1}}{n _{1}}=\frac{2}{100} $
Therefore $ n _{2}-n _{1}=\frac{2}{100}\times n _{1}=\frac{2}{100}\times 392=7.8\approx 8. $
BETA
