Sound waves travel at 350 m/s through warm a | vedclass

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(C) Given: $v_{	ext{air}} = 350 \ m/s$,$v_{	ext{brass}} = 3500 \ m/s$,and frequency $f = 700 \ Hz$.When a sound wave travels from one medium to anot

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increase by a factor of $10$

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(C) Given: $v_{	ext{air}} = 350 \ m/s$,$v_{	ext{brass}} = 3500 \ m/s$,and frequency $f = 700 \ Hz$.When a sound wave travels from one medium to another,its frequency $f$ remains constant.Using the relation $v = f \lambda$,we have $\lambda = \frac{v}{f}$.Since $f$ is constant,the wavelength $\lambda$ is directly proportional to the speed $v$ of the wave $(\lambda \propto v)$.Therefore,$\frac{\lambda_{	ext{brass}}}{\lambda_{	ext{air}}} = \frac{v_{	ext{brass}}}{v_{	ext{air}}}$.Substituting the values: $\frac{\lambda_{	ext{brass}}}{\lambda_{	ext{air}}} = \frac{3500}{350} = 10$.Thus,$\lambda_{	ext{brass}} = 10 \lambda_{	ext{air}}$.The wavelength increases by a factor of $10$.

Column - $A$	Column - $B$
$(a)$ Speed of longitudinal waves in gas and liquid.	$(i)$ $v = \sqrt{\frac{B}{\rho}}$
$(b)$ Speed of longitudinal waves in solid rod.	$(ii)$ $v = \sqrt{\frac{Y}{\rho}}$
$(c)$ Speed of sound in air according to Newton.	$(iii)$ $v = \sqrt{\frac{P}{\rho}}$
$(d)$ Speed of sound in air according to Laplace.	$(iv)$ $v = \sqrt{\frac{\gamma P}{\rho}}$

Sound waves travel at $350 \ m/s$ through warm air and at $3500 \ m/s$ through brass. The wavelength of a $700 \ Hz$ acoustic wave as it enters brass from warm air:

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