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Korvikt [17]
3 years ago
15

The intensity at distance from a spherically symmetric sound source is 100 W/m2. What is the intensity at five times this distan

ce from the source?
Physics
1 answer:
ss7ja [257]3 years ago
3 0

To solve this problem it is necessary to apply the concepts related to intensity as a function of power and area.

Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is

I = \frac{P}{A}

The area of a sphere is given by

A = 4\pi r^2

So replacing we have to

I = \frac{P}{4\pi r^2}

Since the question tells us to find the proportion when

r_1 = 5r_2 \rightarrow \frac{r_2}{r_1} = \frac{1}{5}

So considering the two intensities we have to

I_1 = \frac{P_1}{4\pi r_1^2}

I_2 = \frac{P_2}{4\pi r_2^2}

The ratio between the two intensities would be

\frac{I_1}{I_2} = \frac{ \frac{P_1}{4\pi r_1^2}}{\frac{P_2}{4\pi r_2^2}}

The power does not change therefore it remains constant, which allows summarizing the expression to

\frac{I_1}{I_2}=(\frac{r_2}{r_1})^2

Re-arrange to find I_2

I_2 = I_1 (\frac{r_1}{r_2})^2

I_2 = 100*(\frac{1}{5})^2

I_2 = 4W/m^2

Therefore the intensity at five times this distance from the source is 4W/m^2

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A rope passes over a fixed sheave with both ends hanging straight down. The coefficient of friction between the rope and sheave
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A child has an ear canal that is 1.3 cm long. Assume the speed of sound is v = 344 m/s.
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The  frequencies are (f, f_1) =  (6615.4 \ Hz , 19846.2\ Hz)

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From the question we are told that

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   The  speed of sound is assumed to be  v_s  =  344 \ m/s

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Also the the second harmonic for the pipe (ear canal) is mathematically represented as

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