Question

A 1.2 W point source emits sound waves isotropically. Assuming that the energy of the waves is conserved, find the intensity (a)1.6 m from the source and (b)2.2 m from the source.

Answer 1

Answer:

(a) $I = 3.7\times 10^{-2}\ W/m^{2}$

(b) $I = 1.9\times 10^{-2}\ W/m^{2}$

Explanation:

Given:

Point source $P_{b}=1.2\ W$

Distance from source Part a $L = 1.6\ m$

Distance from source Part b $L = 2.2\ m$

Solution:

Part (a)

Using intensity formula at distance L from an isotropic point.

$I = \frac{P_{b}}{4\pi(L)^{2}}$  -------------------(1)

Substitute $L = 1.6\ m$ and $P_{b}=1.2\ W$ in equation 1..

$I = \frac{1.2}{4\times 3.14(1.6)^{2}}$

$I = \frac{1.2}{32.15}$

$I = 0.037\ W/m^{2}$

$I = 3.7\times 10^{-2}\ W/m^{2}$

Part (b)

Substitute $L = 2.2\ m$ and $P_{b}=1.2\ W$ in equation 1.

$I = \frac{1.2}{4\times 3.14(2.2)^{2}}$

$I = \frac{1.2}{60.79}$

$I = 0.019\ W/m^{2}$

$I = 1.9\times 10^{-2}\ W/m^{2}$

Therefore, the intensity at distance 1.6 m from the source:

$I = 1.9\times 10^{-2}\ W/m^{2}$.

And, the intensity at distance 2.2 m from the source:

$I = 1.9\times 10^{-2}\ W/m^{2}$.