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

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)

    Increased by a factor of 2

  • Option 2)

    Decreased by a factor of 2

  • Option 3)

    Decreased by a factor of 4

  • Option 4)

    unchanged

 

Answers (1)

best_answer

I = 2 \pi ^ {2} a ^ {2} n^{2} v \rho \Rightarrow I \alpha a ^ {2} n^{2}

\Rightarrow \frac{I_{1}}{I_{2}} = \left ( \frac{a_{1}}{a_{2}} \right ) ^ {2} X \left ( \frac{n_{1}}{n_{2}} \right ) ^ {2} = \left ( \frac{1}{2} \right )^{2} X \left ( \frac{1}{1/4} \right )^{2}

\Rightarrow I_{2} = \frac{I_{1}}{4}

 

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.

 

 

 


Option 1)

Increased by a factor of 2

This is incorrect.

Option 2)

Decreased by a factor of 2

This is incorrect.

Option 3)

Decreased by a factor of 4

This is correct.

Option 4)

unchanged

This is incorrect.

Posted by

Aadil

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