A source emits sound uniformly in all directions. A radial line is drawn from this source. On this line, determine the positions of two points, 0.94 m apart, such that the intensity level at one point is 2.12 dB greater than the intensity level at the other. (Enter your answers from smallest to largest.)

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anjalipriya1 anjalipriya1 · Answer 1 · 2020-03-09T05:34:29+01:00

Answer:

Positions of two points are : 3.4 m and 4.3 m.

Explanation:

Let two distance be r and r+0.94.

Now intensity at r, [tex]I_{1\\}[/tex] = [tex]\frac{P}{4 \pi r^{2} }[/tex]

intensity at (r+0.94), [tex]I_{2}[/tex] = [tex]\frac{P}{4 \pi (r+0.94)^{2} }[/tex]

Now for decibel,

[tex]B_{1}[/tex] = 10 log ([tex]\frac{I_{1} }{I_{0} }[/tex])

[tex]B_{2}[/tex] = 10 log ([tex]\frac{I_{2} }{I_{0} }[/tex])

As given on question:

[tex]B_{1}[/tex] - [tex]B_{2}[/tex] = 2.12 dB

⇒ 2.12 dB = 10 log ([tex]\frac{I_{1} }{I_{0} }[/tex]) - 10 log ([tex]\frac{I_{2} }{I_{0} }[/tex])

⇒0.212 = log ([tex]\frac{I_{1} }{I_{2} }[/tex])

⇒ [tex]\frac{I_{1} }{I_{2} }[/tex] = [tex]10^{0.212}[/tex]

⇒ [tex]\frac{(r+0.94)^{2} }{r^{2} }[/tex] = [tex]10^{0.212}[/tex]

⇒ [tex]\frac{r+0.94}{r}[/tex] = [tex]10^{0.106}[/tex]

⇒ r= 3.4 m, r+0.94= 4.3 m

Hence the position of two points is 3.4 m and 4.3 m.