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Physics Question on Gravitation

A point source is emitting sound waves of intensity 16×10−8 Wm−216 \times 10^{-8} \, Wm^{-2} at the origin. The difference in intensity (magnitude only) at two points located at distances of 2 m and 4 m from the origin respectively will be _____ ×10−8 Wm−2\times 10^{-8} \, Wm^{-2}.

Answer

The intensity of sound waves from a point source decreases with the square of the distance from the source. The formula for intensity II at a distance rr from a point source is given by:

I=P4πr2,I = \frac{P}{4\pi r^2}, where PP is the power of the source.

Given Values: Intensity at the origin I0=16×10−8 Wm−2I_0 = 16 \times 10^{-8} \, Wm^{-2}. Distances: r1=2 mr_1 = 2 \, m and r2=4 mr_2 = 4 \, m.

Intensity at Distances r1r_1 and r2r_2: The intensity at distance r1=2 mr_1 = 2 \, m:

I1=I0(r0r1)2=16×10−8×(12)2=16×10−8×14=4×10−8 Wm−2.I_1 = I_0 \left( \frac{r_0}{r_1} \right)^2 = 16 \times 10^{-8} \times \left( \frac{1}{2} \right)^2 = 16 \times 10^{-8} \times \frac{1}{4} = 4 \times 10^{-8} \, Wm^{-2}.

The intensity at distance r2=4 mr_2 = 4 \, m:

I2=I0(r0r2)2=16×10−8×(14)2=16×10−8×116=1×10−8 Wm−2.I_2 = I_0 \left( \frac{r_0}{r_2} \right)^2 = 16 \times 10^{-8} \times \left( \frac{1}{4} \right)^2 = 16 \times 10^{-8} \times \frac{1}{16} = 1 \times 10^{-8} \, Wm^{-2}.

Calculating the Difference in Intensity: The difference in intensity ΔI\Delta I between the two points:

ΔI=I1−I2=(4×10−8−1×10−8) Wm−2=3×10−8 Wm−2.\Delta I = I_1 - I_2 = (4 \times 10^{-8} - 1 \times 10^{-8}) \, Wm^{-2} = 3 \times 10^{-8} \, Wm^{-2}.