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Question: Answered & Verified by Expert
A stationary point source of sound emits sound uniformly in all directions in a non-absorbing medium. Two points $P$ and $Q$ are at a distance of $4 \mathrm{~m}$ and $9 \mathrm{~m}$ respectively from the source. The ratio of amplitudes of the waves at $P$ and $Q$ is
PhysicsWaves and SoundKCETKCET 2009
Options:
  • A $\frac{3}{2}$
  • B $\frac{4}{9}$
  • C $\frac{2}{3}$
  • D $\frac{9}{4}$
Solution:
2664 Upvotes Verified Answer
The correct answer is: $\frac{9}{4}$



For an isotropic point source of power P, intensity I at a distance $\mathrm{r}$ from it is
$$
\mathrm{I}=\frac{\mathrm{P}}{4 \pi \mathrm{r}^{2}}
$$
Since power P remains the same,
$$
\therefore \quad \frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{2}=\left(\frac{9}{4}\right)^{2}
$$
$\because \quad I \propto \mathrm{A}^{2}$
where $A$ is the amplitude of a wave $\therefore \quad \frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\sqrt{\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}}=\frac{9}{4}$

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