Question

Wavelengths of two nodes in air are $\dfrac{90}{175}m$ and $\dfrac{90}{173}m$. Each node produces four beats per second with a third node of fixed frequency. The node of higher frequency will be
A. 688Hz
B. 692Hz
C. 700Hz
D. 708Hz

Answer 1

Hint: You could find which among the given two nodes will have higher frequency from the given value of wavelength. You could find the relation between these two frequencies from the information given on the beats produced by them with the third node. Now substitute in terms of velocity of sound and find the value of the same. Dividing the velocity by wavelength will give the required frequency.
Formula used:
Frequency,
$f=\dfrac{v}{\lambda }$

Complete answer:
In the question, we are given the wavelengths of two nodes in air as,
${{\lambda }_{1}}=\dfrac{90}{175}m$
${{\lambda }_{2}}=\dfrac{90}{173}m$
Clearly, by comparing the denominators we find that, ${{\lambda }_{1}}\langle {{\lambda }_{2}}$ which would directly imply that their respective frequencies ${{f}_{1}}$ and ${{f}_{2}}$ would be related as, ${{f}_{1}}\rangle {{f}_{2}}$. Let the third mentioned frequency be f. we said that each of the given two nodes produces 4 beats per second with this third node. That is,
${{f}_{1}}-f=4$ ……………………………………… (1)
And,
$f-{{f}_{2}}=4$ ………………………………………. (2)
Adding equations (1) and (2), we get,
${{f}_{1}}-{{f}_{2}}=8$ ………………………………………. (3)
We know that the frequency is related to velocity of sound (v) and wavelength$\left( \lambda \right)$ as,
$f=\dfrac{v}{\lambda }$ ……………………………………………….. (4)
So equation (3) now becomes,
$\dfrac{v}{{{\lambda }_{1}}}-\dfrac{v}{{{\lambda }_{2}}}=8$
$\Rightarrow v\left( \dfrac{1}{{{\lambda }_{1}}}-\dfrac{1}{{{\lambda }_{2}}} \right)=8$
$\Rightarrow v\left( \dfrac{175}{90}-\dfrac{173}{90} \right)=8$
$\Rightarrow v\left( \dfrac{2}{90} \right)=8$
$\therefore v=360m{{s}^{-1}}$ ……………………………………………….. (5)
We have already concluded that ${{f}_{1}}$ is higher among the two nodes given, so the equation (4) for ${{f}_{1}}$ would be,
${{f}_{1}}=\dfrac{v}{{{\lambda }_{1}}}$
Substituting (5),
${{f}_{1}}=\dfrac{360}{\left( \dfrac{90}{175} \right)}$
$\therefore {{f}_{1}}=700Hz$
Therefore, we found the node of higher frequency to have a frequency of 700Hz.

Hence, option C is found to be the correct answer.

node:
We should understand that beat frequency can be given by the difference in the frequencies of the nodes produced by some device that interferes to produce the beats. Beats can also be defined as the periodically repeating fluctuations in the intensity of the sound heard at the end of this interference. No sound can be heard when destructive interference occurs.