If the amplitude of sound is doubled and the frequency reduced to one-fourth, the intensity of sound at the same point will be

  • Option 1)

    Increasing by a factor of 2

  • Option 2)

    Decreasing by a factor of 2

  • Option 3)

    Decreasing by a factor of 4

  • Option 4)

    Unchanged

 

Answers (1)

As we learnt in 

Intensity of the wave -

I = \frac{1}{2}\rho \omega ^{2}A^{2}V

\rho = mass density

\omega = angular frequency

A = Amplitude

V = Wave speed

 

- wherein

flow of energy per unit area of cross section of the string in the unit time is known as intensity of wave.

 

 Intensity of sound \propto \:A^{2}\nu^{2}

\therefore\:\frac{I_{2}}{I_{1}}=(\frac{A_{2}}{A_{1}})^{2}.(\frac{\nu_{2}}{\nu_{1}})^{2}=4\times \frac{1}{16}=\frac{1}{4}

 


Option 1)

Increasing by a factor of 2

This option is incorrect.

Option 2)

Decreasing by a factor of 2

This option is incorrect.

Option 3)

Decreasing by a factor of 4

This option is correct.

Option 4)

Unchanged

This option is incorrect.

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