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meriva
3 years ago
9

By what factor must the amplitude of a sound wave be decreased in order to decrease the intensity by a factor of 3?

Physics
1 answer:
s344n2d4d5 [400]3 years ago
4 0

Answer:

Amplitude is decreased by a factor of \sqrt3 if intensity is decreased by a factor of 3.

Explanation:

Intensity of a sound wave is directly proportional to the square of its amplitude.

Therefore, if intensity is I and amplitude is A, then

I=kA^2, where, k is constant of proportionality.

Now, if intensity of sound wave is decreased by a factor of 3. So,

New intensity is, I_{new}=\frac{I}{3}

I_{new}=kA_{new}^2\\\frac{I}{3}=kA_{new}^2

Plug in kA^2 for I. This gives,

\frac{kA^2}{3}=kA_{new}^2\\A_{new}^2=\frac{A^2}{3}\\A_{new}=\sqrt{\frac{A^2}{3}}=\frac{A}{\sqrt{3}}

Therefore, amplitude is decreased by a factor of \sqrt3.

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a)   I2 = 3 (o-10) / (o- 30) , b)   h ’/h=  3 (o-10) / o (o-30)

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The builder's equation is

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Where f is the focal length, or e i are the distance to the object and image, respectively

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         1 / i1 = 1 / f - 1 / o

         1 / i1 = 1/15 - 1 / o

         i1 = o 15 / (o-15)

Let's calculate the distance to the image of the second lens, for this the image of the first is the distance to the object of the second

        o2 = d-i1

We write the builder equation

       1 / f2 = 1 / o2 + 1 / i2

       1 / i2 = 1 / f2 -1 / o2

       1 / i2 = 1 / f2 - 1 / (d-i1)

       1 / i2 = 1/20 - 1 / (d-i1)            (1)

Let's evaluate the last term

      d-i1 = d - 15 o / (o-15)

      d-i1 = (d (o-15) - 15 o) / (o-15)

      d- i1 = (30 or -30 15 -15 o) / (o-15)

      d-i1 = (15 or - 450) / (o- 15)

      d-i1 = = (15 or -450) / (o-15)

replace in 1

      1 / i2 = 1/20 - (or - 15) / (15 or -450)

      1 / i2 = [(15 o-450) - (o-15) 20] / (15 or -150)

      1 / i2 = (15 or - 450 - 20 or + 300) / (15 or - 150)

      1 / i2 = (-5 or -150) / (15 or -150)

      1 / i2 = (or -30) / (3 or - 30)

      I2 = 3 (o-10) / (o- 30)

Part B

The height of the image, we use the magnification equation

     m = h ’/ h = - i / o

     h ’= - h i / o

In our case

     h ’= h i2 / o

     h ’= h 3 (o-10) / o (o-30)

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