The wavelengths of two notes in air are frac{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the frequency of the third note be $n$ and the velocity of sound be $v$.The frequencies of the two notes are $n_1 = \frac{v}{\lambda_1} = \fra

Vedclass · Accepted Answer

$360$

Vedclass · Answer

(D) Let the frequency of the third note be $n$ and the velocity of sound be $v$.The frequencies of the two notes are $n_1 = \frac{v}{\lambda_1} = \frac{v}{36/195} = \frac{195v}{36}$ and $n_2 = \frac{v}{\lambda_2} = \frac{v}{36/193} = \frac{193v}{36}$.Since each note produces $10$ beats per second with the third note, we have:$\frac{195v}{36} - n = 10$  ---$(i)$$n - \frac{193v}{36} = 10$  ---(ii)Adding equations $(i)$ and (ii):$\frac{195v}{36} - \frac{193v}{36} = 10 + 10$$\frac{2v}{36} = 20$$\frac{v}{18} = 20$$v = 360 \,m/s$.

The wavelengths of two notes in air are $\frac{36}{195} \,m$ and $\frac{36}{193} \,m$. Each note produces $10$ beats per second separately with a third note of fixed frequency. The velocity of sound in air in $m/s$ is

Similar Questions

The disc of a siren containing $60$ holes rotates at a constant speed of $360\,rpm$. The emitted sound is in unison with a tuning fork of frequency .... $Hz$

The wavelength of one wave is $99 \ cm$ and that of another is $100 \ cm$. If the speed of sound is $396 \ m/s$,the number of beats heard per second is:

The frequency of a tuning fork is $n \ Hz$ and the velocity of sound in air is $V \ m/s$. When the tuning fork completes $x$ vibrations,the distance traveled by the wave is:

In Melde's experiment in the transverse mode,the frequency of the tuning fork and the frequency of the waves in the strings are in the ratio

Two sound waves of slightly different frequencies propagating in the same direction produce beats due to

Vedclass Test Series

Exam Paper Generator

Online Exam Module