Question

A sound source of frequency 512 Hz is producing 6 beats with a guitar. If the string of the guitar is stretched slightly then beat frequency decreases. The original frequency of guitar is
A. 506 Hz
B. 512 Hz
C. 518 Hz
D. 524 Hz

Answer 1

Hint:In this question, we need to determine the original frequency of the guitar such that a sound source was initially producing 6 beats with it. For this we will use the relation between the frequency of the sound source, frequency of the guitar and the number of the beats as ${n_s} \sim {n_g} = {x_b}$ where, ${n_s}$ and ${n_g}$ are the frequencies of the sound source and the guitar respectively and ${x_b}$ is the beats of the guitar.

Complete step by step answer:
Sound source frequency \[{n_s} = 512Hz\]
Let the frequency of guitar be \[{n_g}\]which is to be calculated
We know that the beats are produced in instruments when there is a frequency difference, so let us consider guitar produces ‘x’ numbers of beats and ‘x’ number of beats can be produced by
\[{n_s} - {n_g} = x - - (i)\]
\[{n_g} - {n_s} = x - - (ii)\]
Now it is given that the string of the guitar is stretched so it will create tension on the string which is given by the equation
\[\sqrt T \propto {n_g} - - (iii)\]
Since the string of the guitar is stretched so the tension in the string and also frequency of the guitar increases, now let us observe the equation (i) where the source frequency \[{n_s} = 512Hz\]is constant, now when the frequency of the guitar increases the number of beats decreases and this is what the question desires hence case (i) satisfies, no by solving
\[
  {n_s} - {n_g} = x \\
  \Rightarrow 512 - {n_g} = 6 \\
  \Rightarrow {n_g} = 512 - 6 \\
   = 506Hz \\
 \]
Therefore the original frequency of guitar is \[ = 506Hz\]
Option A is correct.

Note:It is interesting to note here that, many a times the frequency of the sound source or the guitar can be greater than the other and so we need to use the formula ${n_s} \sim {n_g} = {x_b}$ accordingly.