Question 83: Waves

Question

A point source is emitting sound waves of intensity $$16 	imes 10^{-8} \mathrm{~Wm}^{-2}$$ at the origin. The difference in intensity (magnitude only) at two points located at a distances of $$2 m$$ and $$4 m$$ from the origin respectively will be _________ $$	imes 10^{-8} \mathrm{~Wm}^{-2}$$.

Accepted Answer

To solve this problem, we need to understand the relationship between the intensity of sound waves and the distance from the source. The intensity $I$ of sound waves from a point source decreases with the square of the distance $r$ from the source, according to the inverse square law, which can be expressed as:
$$ I \propto \frac{1}{r^2} $$
This means that if the distance is doubled, the intensity becomes one-fourth of its initial value because $ (2r)^2 = 4r^2 $.
The initial intensity given at the origin (source) is:
$$ I_0 = 16 	imes 10^{-8} \mathrm{~Wm}^{-2} $$
Let's call $ I_1 $ the intensity at $ r = 2 	ext{ m} $ and $ I_2 $ the intensity at $ r = 4 	ext{ m} $. Using the inverse square law, we can write:
$$ I_1 = \frac{I_0}{(2)^2} = \frac{I_0}{4} $$
and
$$ I_2 = \frac{I_0}{(4)^2} = \frac{I_0}{16} $$
Now substitute the given value for $ I_0 $ to find $ I_1 $ and $ I_2 $:
$$ I_1 = \frac{16 	imes 10^{-8}}{4} = 4 	imes 10^{-8} \mathrm{~Wm}^{-2} $$
$$ I_2 = \frac{16 	imes 10^{-8}}{16} = 1 	imes 10^{-8} \mathrm{~Wm}^{-2} $$
Now to find the difference in intensity (magnitude only) between the two points, we subtract $ I_2 $ from $ I_1 $:
$$ \Delta I = | I_1 - I_2 | $$
$$ \Delta I = | 4 	imes 10^{-8} - 1 	imes 10^{-8} | $$
$$ \Delta I = 3 	imes 10^{-8} \mathrm{~Wm}^{-2} $$
So the difference in intensity (magnitude only) at the two points is:
$$ 3 	imes 10^{-8} \mathrm{~Wm}^{-2} $$

Waves

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

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