Name: Power Transmitted by a Wave on a String
Uploaded: 2019-07-19
Duration: 7 min 4 s
Description: We know that a wave transfers energy from one point to another, but to find the power transmitted by a wave on a string, watch this video.

Question

A point isotropic sound source is located on the perpendicular to the plane of a ring, drawn through the centre

O

of the ring. The distance between the point

O

and the source is

l = 1.00

m

and the radius of the ring is

R = 0.50

m

. Find the mean energy flow across the area enclosed by the ring, if at point

O

, the intensity of sound is equal to

I_{0} = 30

μW m^{- 2}

. The damping of the waves is negligible.

Accepted Answer

For the point source given in the problem, the intensity at point $O$ is $I_{0}$ . This means the power of the point source will be, $P_{0} = 4 π l^{2} I_{0}$ . At any general point, the intensity would be, $I = \frac{P_{0}}{4 π a^{2}}$ . The flow of energy per unit time, through a small area $d A$ , is

$d E = I (d A) cosθ$ , where $θ$ is the angle between the area vector and the direction of propagation of wave. This equation is similar to the calculation of flux of electric field. In fact, the energy crossing an area per unit time is the flux of intensity of a wave.

$d E = (\frac{P_{0}}{4 π a^{2}}) (d A) (cosθ)$ . Now, $\frac{d A cosθ}{a^{2}} = d Ω$ , the infinitesimal solid angle subtended by the small area at point $P$ .

As, shown in the figure below.

$d E = \frac{P_{0}}{4 π} d Ω$ $\Rightarrow E = \frac{P_{0}}{4 π} Ω$ , where $O$ is the total solid angle subtended by the ring at point $l = 1.00$ . For a ring that subtends a half-angle $α$ at a point, the solid angle is, $Ω = 2 π (1 - \cos α) = 2 π (1 - \frac{l}{\sqrt{l^{2} + R^{2}}})$ $\Rightarrow E = \frac{P_{0}}{4 π} \times 2 π (1 - \frac{l}{\sqrt{r^{2} + R^{2}}})$

$\Rightarrow E = \frac{4 π I_{0} l^{2}}{4 π} \times 2 π (1 - \frac{l}{\sqrt{r^{2} + R^{2}}})$ $\Rightarrow E = 2 π I_{0} l^{2} (1 - \frac{l}{\sqrt{r^{2} + R^{2}}})$ $\Rightarrow E = 1.99 μW$ .

Important Questions on Wave Motion on a String

Learn and Prepare for any exam you want !