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The average power transmitted across a c...

The average power transmitted across a cross section by two sound waves moving in the same direction are equal. The wavelength of two sound waves are in the ratio of `1:2`, then find the ratio of their pressure amplitudes.

Text Solution

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The correct Answer is:
1

`P=(1)/(2)rho omega^(2)A^(2)sV`
Since `(lamda_(1))/(lamda)=(1)/(2),(f_(1))/(f_(2))=(omega_(1))/(omega_(2))=(2)/(1)`
Since `P_(1)=P_(2),omega_(1)A_(1)=omega_(2)A_(2), (A_(1))/(A_(2))=(omega_(1))/(omega_(2))=(1)/(2)`
Pressure amplitude `P_(0)=B_(0)Ak`
`(P_(0))_(1)(P_(0))_(2)=((A_(1))/(A_(2)))((k_(1))/(k_(2)))=((A_(1))/(A_(2)))((lamda_(2))/(lamda_(1)))`
`=((1)/(2))((2)/(1))=1`
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