Two sound waves of wavelengths lambda_1 and l | vedclass

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

Question

(A) In the same medium,the speed of sound $v$ is constant.The relationship between speed,frequency $(f)$,and wavelength $(\lambda)$ is given by $v = f

Vedclass · Accepted Answer

$\frac{n\lambda_1\lambda_2}{\lambda_2 - \lambda_1}$

Vedclass · Answer

(A) In the same medium,the speed of sound $v$ is constant.The relationship between speed,frequency $(f)$,and wavelength $(\lambda)$ is given by $v = f\lambda$,which implies $f = \frac{v}{\lambda}$.Let $f_1$ and $f_2$ be the frequencies of the two waves. Then $f_1 = \frac{v}{\lambda_1}$ and $f_2 = \frac{v}{\lambda_2}$.Since $\lambda_2 > \lambda_1$,it follows that $f_1 > f_2$.The number of beats per second is the difference in frequencies: $n = f_1 - f_2$.Substituting the expressions for $f_1$ and $f_2$:$n = \frac{v}{\lambda_1} - \frac{v}{\lambda_2}$$n = v \left( \frac{\lambda_2 - \lambda_1}{\lambda_1\lambda_2} \right)$Solving for $v$:$v = \frac{n\lambda_1\lambda_2}{\lambda_2 - \lambda_1}$

Two sound waves of wavelengths $\lambda_1$ and $\lambda_2$ $(\lambda_2 > \lambda_1)$ produce $n$ beats per second. The speed of sound is:

Similar Questions

Two sound waves with wavelengths $5.0\, m$ and $5.5\, m$ respectively,each propagate in a gas with velocity $330\, m/s$. We expect the following number of beats per second.

When two waves of almost equal frequencies $v_1$ and $v_2$ reach a point simultaneously,the time interval between successive maxima is

When a tuning fork is vibrating,the vibrations of the two prongs

The wavelengths of two sound notes in air are $\frac{40}{195} \,m$ and $\frac{40}{193} \,m$. Each note produces $9$ beats per second separately with a third note of fixed frequency. The velocity of sound in air in $m/s$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module