General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Chapter 22, Problem 56E
To determine

The intensity level heard at 50m.

Expert Solution & Answer
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Answer to Problem 56E

The intensity level heard at 50m is 56.02dB_.

Explanation of Solution

Write the expression for intensity level of sound.

    β=10log10(I1I0)        (I)

Here, β is the intensity level of sound, I0 is the lowest audible intensity and I1 is the intensity of sound at 10m distance.

Given that the intensity level of sound is 70dB at a distance of 10m.

Apply the above condition in equation (I), and rearrange to find I1.

    70dB=10log10(I1I0)7=log10(I1I0)107=(I1I0)I1=107×I0        (II)

Since the lowest audible intensity is equal to 1012Wm2, equation (II) is written as

    I1=(107)(1012Wm2)=105Wm2

Write the relation between acoustic power in terms of intensity.

    P=I1(4πr2)        (III)

Here, P is the acoustic power at 10m distance, and r is the distance from the source.

Since the intensity of sound at 10m distance is 105Wm2.

Use the above condition in equation (III).

    P=(105Wm2)(4π(10m)2)=0.01256W        (IV)

Write the acoustic power at 50m in terms of intensity.

    P=I(4πr2)        (V)

Here, P is the acoustic power at 50m distance, I is the intensity at 50m, and r is the distance from the source

Rearrange the above equation to find I.

    I=P4πr2

Since the acoustic power at 50m distance is 0.01256W, and the distance from the source is 50m the above equation is written as

    I=0.01256W4π(50m)2=4×107Wm2        (VI)

Write the expression for intensity level of sound at a distance of 50m.

    β=10log10(II0)        (VII)

Here, β is the intensity level of sound, I0 is the lowest audible intensity and I is the intensity of sound at a distance of 50m.

Conclusion:

Substitute 4×107Wm2 for I, and 1012Wm2 for I0 in equation (VII), to find β.

    β=10log10(4×107Wm21012Wm2)=56.02dB

Therefore, the intensity level heard at 50m is 56.02dB_.

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