Two sound waves of wavelengths lambda_1 and l | vedclass

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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:

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