JEE Main On 16 April 2018 Question 23
Question: Two sitar strings, A and B, playing the note ‘Dha’ are slightly out of tune and produce beats and frequency $ 5,Hz. $ The tension of the string B is slightly increased and the beat frequency is found to decrease by $ 3,Hz. $ If the frequency of is , the original frequency of A is 425 Hz, the original frequency of B is [JEE Main 16-4-2018]
Options:
A) 430 Hz
B) 428 Hz
C) 422 Hz
D) 420 Hz
Show Answer
Answer:
Correct Answer: A
Solution:
The difference in frequency is known as the number of beats.
Here, frequency of $ A,f _{A}=324,Hz $ We know,
Frequency of $ B,f _{B}=f _{A},\pm $ beat frequency $ =425\pm 5 $ $ =420Hzor430Hz $ Now,
if tension in string slightly reduced then its frequency also reduce from $ f _{B} $
Now, if tension in the string is slightly reduced its frequency will also reduce from 324 Hz.
Now, if $ f _{B}=420 $ reduces, then beat frequency should increase which is not the case
but if $ f _{B}=430,Hz $ then beat frequency should decrease, which is the case hence =430 Hz.
$ f _{B}=430,Hz. $ Let the frequency of string $ B= $ $ f _{B} $ and frequency of string
Initially beat frequency = $ =5Hz $ Now the tension in string B is increased
so frequency b will decrease as frequency is inversely proportional to tension.
$ f _{A}=425,HZ $ $ f _{B} $ can be either 430HZ or 420HZ
But when tension is increased frequency $ f _{B} $ will decrease and it is given that
it produces beat frequency $ =3HZ $ Which is only possible when $ fB=230Hz $ $ F _{B}=430HZ $
