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 A is $425 Hz$, the original frequency of B is
A. $430 Hz$
B. $428 Hz$
C. $422 Hz$
D. $420 Hz$

Answer 1

Hint: Beat is defined as the modulus of difference between the frequencies of two sound waves. Therefore, frequency of B $ = $ frequency of A$ \pm $ beat frequency. The higher the tension, the higher the frequency of the fundamental. Now on observing the new beat frequencies, the frequency of B which gives the frequency after change greater than the original frequency is the answer.

Complete answer:
Given, the frequency of A, ${f_A} = 425 Hz$.
The beat frequency, $f = 5Hz.$
Let the frequency of B be ${f_B}$.
We know that beat frequency is defined as the difference in frequencies of the two sound sources. Therefore we get the frequency of B,
${f_B} = {f_A} \pm f$.
$ \Rightarrow {f_B} = 425 \pm 5$
Therefore frequency of B is either $430\, or\, 420Hz$.

We know that the relation between frequency and tension is given as
$\nu = \dfrac{n}{{2L}}\sqrt {\dfrac{T}{\mu }} $, therefore frequency of B should increase with increase in tension.
New frequency of B is, ${f_B} = 425 \pm 3 = 428\,or\,428$ Hz.
Therefore frequency of B increases corresponding to 420Hz original frequency of B.

Therefore the original frequency of B is 420Hz.

Note: One should know that beat is defined as the modulus of difference in frequencies of two sources. On removing the modulus sign one should remember to use the plus-minus sign. Frequency increases with the increase in tension.