A point source emits sound equally in all directions in a non-absorbing medium. Two points P and Q a

Q: Why does the amplitude decrease with distance if the medium is non-absorbing?

Even though no energy is lost to the medium, the total energy of the wavefront is spreading over an ever-increasing spherical surface area ($4\pi r^2$). To keep the total energy conserved, the energy per unit area (intensity) must decrease, which causes the amplitude of particle vibration to drop as $1/r$.

Q: What would change if the source were a line source instead of a point source?

For a line source (like a long vibrating cylinder), the wavefronts are cylindrical. The surface area of a cylinder scales as $2\pi r L$, so the intensity decreases as $1/r$. Consequently, the amplitude decreases as $1/\sqrt{r}$ instead of $1/r$.

Q: Does this inverse relationship hold true for plane waves?

No. For an ideal plane wave, the wavefronts do not spread out; they remain parallel planes. Therefore, the intensity and amplitude of an ideal plane wave remain constant with distance in a non-absorbing medium.

Q: What happens to the amplitude if the medium actually absorbs sound?

If the medium is absorbing, energy is lost as heat. The amplitude will decay even faster, following an exponential decay superimposed on the geometric spreading: $A(r) \propto \frac{e^{-\alpha r}}{r}$, where $\alpha$ is the absorption coefficient.

Q: Is the ratio of intensities the same as the ratio of amplitudes?

No. Since intensity is proportional to the square of amplitude ($I \propto A^2$), the ratio of intensities at $P$ and $Q$ would be the square of the amplitude ratio, which is $\left(\frac{25}{9}\right)^2 = \frac{625}{81}$.

Question

A point source emits sound equally in all directions in a non-absorbing medium. Two points $P$ and $Q$ are at a distance $9	ext{ m}$ and $25	ext{ m}$ respectively from the source. The ratio of amplitudes of the waves at $P$ and $Q$ is ……

Sigma Percentile · Accepted Answer

### The Magic of Point Sources

Imagine standing in an open, quiet field. A single firecracker pops in the distance. The sound doesn't just travel in a straight line toward you; it expands as an ever-growing sphere of energy, rushing to fill the three-dimensional space around it.

This is the essence of a **point source** of sound. Because the medium is non-absorbing, no energy is lost to heat or friction. The total power, $P_0$, emitted by the source remains completely conserved as it propagates outward.

But if the energy is conserved, why does the sound get quieter as you move away? This is where geometry plays its beautiful role.

### The Inverse Square Law of Intensity

As the sound wave travels a distance $r$, the initial power $P_0$ must spread itself over the surface of a sphere of radius $r$.

The surface area of this sphere is given by the classic geometric formula:

$$A_{	ext{sphere}} = 4\pi r^2$$

Since intensity $I$ is defined as the power flowing per unit area, we can write:

$$I = \frac{P_0}{4\pi r^2}$$

Because $P_0$ and $4\pi$ are constants, we discover that the intensity is inversely proportional to the square of the distance:

$$I \propto \frac{1}{r^2}$$

This is the famous **Inverse Square Law**. If you double your distance from the source, the sound intensity drops to one-fourth of its original value!

### Connecting Intensity to Amplitude

But the question doesn't ask about intensity; it asks about **amplitude**, the maximum displacement of the air particles as they vibrate.

How do intensity and amplitude relate? From the physics of wave motion, the energy carried by a wave—and thus its intensity—is directly proportional to the square of its amplitude:

$$I \propto A^2$$

This makes intuitive sense: a wave with twice the amplitude requires four times the energy to create.

Now, let's combine our two proportionalities:

$$A^2 \propto I \propto \frac{1}{r^2}$$

Taking the square root of both sides, we get a wonderfully simple relationship:

$$A \propto \frac{1}{r}$$

For a spherical wave, the amplitude of vibration decreases inversely with the distance $r$. This is a crucial distinction from plane waves, where amplitude remains constant over distance!

### Calculating the Ratio

We are given two points, $P$ and $Q$, at distances $r_P = 9	ext{ m}$ and $r_Q = 25	ext{ m}$ from the source.

Using our inverse relationship, we can set up the ratio of their amplitudes:

$$\frac{A_P}{A_Q} = \frac{r_Q}{r_P}$$

Now, we simply substitute the given values:

$$\frac{A_P}{A_Q} = \frac{25}{9}$$

Thus, the ratio of the amplitudes of the waves at $P$ and $Q$ is **$\frac{25}{9}$**.

This elegant result shows that the amplitude at $P$ (which is closer) is nearly three times larger than the amplitude at $Q$ (which is further away).

Animated Solution for Physics - Waves: A point source emits sound equally in all directions in a non-absorbing medium. Two points $P$ and $Q$ are at a distance $9 m$ and $25 m$ respectively from the source. The ratio of amplitudes of the waves at $P$ and $Q$ is ……

Visualized Solution

Visualizing the Physical Setup

The Inverse Square Law of Intensity

Relating Intensity to Amplitude

Combining the Proportionalities

Setting Up the Ratio

Substituting the Given Values

Final Conclusion

The Way Forward

The Sigma Insight: Wave Equation and Wave Speed

The Magic of Point Sources

The Inverse Square Law of Intensity

Connecting Intensity to Amplitude

Calculating the Ratio

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Visualized Solution

Visualizing the Physical Setup

The Inverse Square Law of Intensity

Relating Intensity to Amplitude

Combining the Proportionalities

Setting Up the Ratio

Substituting the Given Values

Final Conclusion

The Way Forward

The Sigma Insight: Wave Equation and Wave Speed

The Magic of Point Sources

The Inverse Square Law of Intensity

Connecting Intensity to Amplitude

Calculating the Ratio

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A small speaker delivers 2 W of audio output. At what distance from the speaker will one detect 120 dB intensity sound? [Take, reference intensity of sound as 10−12 W/m2]

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A small speaker delivers $2 W$ of audio output. At what distance from the speaker will one detect $120 dB$ intensity sound? [Take, reference intensity of sound as $1 0^{- 12} W/m^{2}$ ]