i have two physics questions relating to intensity of wave?

Question

a small source of sound radiates energy equally in all directions at a particular frequency the intensity of sound 1.0m from the source is 1 raise to power -5 W/M2 corresponding to an amplitude of oscillation of the air molecules of 70 micrometer.assuming sound is propagated without energy loss
what will be
the intensity of sound
the amplitude of oscillation of the air molecule at a distance of 5.0m from the source.
Q2
a point source of sound radiates energy uniformally in all directions .at a distance of 3m from the source the amplitude of vibration of air molecule is 1 raise to power -7.assuming no energy is absorbed .calculate the amplitude of vibration 5m from the source
answer:
Q1:a:4 raise to power -7 W/M2
b:14micrometer
Q2 6 raise to power -8 m
plz solve these i could not solve them

peter,70 is in micrometer and 5 in meter when we divide h answer can come in micrometer,plz explain it again,whole question if u h time

Answer 1

The sound intensity due to a point source radiating equally in all directions (spherically symmetric case) is proportional to 1/r^2, where r is the distance from the source of the sound:

I = k/r^2, where k is a constant of proportionality.

If we know the intensity (I1) at one distance (r1), then the intensity (I2) at a second distance (r2) is given by:

I2 = I1*(r1/r2)^2

In Q1, we are told that the sound intensity at 1 m is 10^-5 W/m^2, and we want to know the intensity at 5 m. Plugging these values into the above equation, we have:

I2 = 10^-5 W/m^2 * (1m/5m)^2 = (10^-5 W/m^2)/25 = 4*10^-7 W/m^2

The intensity of a sound wave is proportional to the square of the amplitude of the displacement of a particle in the medium through which the sound is propagating:

I = m*x^2,

where m is a constant of proportionality, and x is the amplitude of the displacement. Plugging this into the equation above that relates intensity and distance, we have that:

(x2)^2 = (x1)^2 * (r1/r2)^2
x2 = x1*r1/r2

The amplitude of the displacement is inversely proportional to 1/r.

For Q1 we are told that x1 = 70 microns at 1 meter, and are asked what x2 is at 5 meters. Plugging these values into the above equation, we have:

x2 = 70 microns * 1/5 = 14 microns

For Q2, we are told that x1 = 10^-7 m at r1 = 3 m, and are asked what the amplitude is at 5 m. Again plugging these into the above equation, we have:

X2 = 10^-7 m * (3/5) = 6*10^-8 m

Answer 2

Salaam,
by definition, 0 dB is 10^-3 watts per square metre, so since the decibel scale is a logarithmic one in which 10 decibels is 10 times a 10 fold increase, 10^-5 watts/M2 is-20 dB. That's the intensity. Sound obeys an inverse square law, and intensity is proportional to the square of the amplitude of oscillation, so the amplitude is inversely proportional to the distance from the source. So 5 metres from the source, the amplitude will be 70/5=14 micrometres. Q2;3/5 times 10^-7 = 6 times 10^-8.

Answer 3

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