Two coherent sources of sound S1 and S2, produce sound waves of the same wavelength lambda = 1 m, in

Q: Why does the path difference increase when moving away from $S_1$?

Since the distance to $S_2$ is fixed at 2 meters, moving away from $S_1$ means the distance $S_1L'$ is increasing. The path difference is defined as $S_1L' - S_2L'$. Because $S_1L'$ is getting larger and $S_2L'$ is constant, the overall path difference must increase.

Q: How do we know the next maximum is exactly at $\Delta x = \lambda$?

The initial state is a minimum at a path difference of $\frac{\lambda}{2}$. Because the listener is moving in a way that increases the path difference, the very next condition for constructive interference (a maximum) is the next integer multiple of $\lambda$, which is $1\lambda$.

Q: What if the listener moved *towards* $S_1$ instead?

If the listener moved towards $S_1$ along the same arc, the distance $d$ would decrease, meaning the path difference would decrease from $\frac{\lambda}{2}$. The adjacent maximum in that direction would occur when the path difference drops to $0$.

Q: Why is the path difference $S_1L - S_2L$ and not the other way around?

Path difference is the absolute difference in distances. Since $S_1L$ is the hypotenuse of the right-angled triangle, it is inherently longer than the leg $S_2L$. We subtract the smaller distance from the larger one to keep the path difference positive.

Q: Does the initial phase of the sources matter?

Absolutely! The problem explicitly states they are 'in phase'. If the sources were initially out of phase by $\pi$ radians, the conditions for maxima and minima would be completely swapped.

Question

Two coherent sources of sound $S_1$ and $S_2$, produce sound waves of the same wavelength $\lambda = 1	ext{ m}$, in phase. $S_1$ and $S_2$ are placed 1.5 m apart (see figure). A listener, located at $L$, directly in front of $S_2$ finds that the intensity is at a minimum when he is 2 m away from $S_2$. The listener moves away from $S_1$, keeping his distance from $S_2$ fixed. The adjacent maximum of intensity is observed when the listener is at a distance $d$ from $S_1$. Then, $d$ is

Options:

(A) 12 m

(B) 5 m

(C) 2 m

(D) 3 m

Sigma Percentile · Accepted Answer

### Visualizing the Setup

Imagine you are standing in an open field with two massive speakers, $S_1$ and $S_2$, placed exactly $1.5	ext{ m}$ apart. These speakers are perfectly synchronized, emitting sound waves with a wavelength of $\lambda = 1	ext{ m}$. You start your journey at a specific point $L$, which is located directly in front of speaker $S_2$ at a distance of $2	ext{ m}$.

To understand what you hear at point $L$, we must analyze the **path difference**—the difference in the distance the sound travels from each speaker to reach your ears. The sound from $S_2$ travels a straightforward $2	ext{ m}$. But what about the sound from $S_1$?

### The Initial State (Minima)

Since $L$ is directly in front of $S_2$, the points $S_1$, $S_2$, and $L$ form a perfect right-angled triangle. We can use the Pythagorean theorem to find the distance from $S_1$ to $L$:

$$S_1L = \sqrt{S_1S_2^2 + S_2L^2}$$

Substituting the given values:

$$S_1L = \sqrt{1.5^2 + 2^2} = \sqrt{2.25 + 4} = \sqrt{6.25} = 2.5	ext{ m}$$

Now, we calculate the initial path difference, $\Delta x_1$, at point $L$:

$$\Delta x_1 = S_1L - S_2L = 2.5	ext{ m} - 2.0	ext{ m} = 0.5	ext{ m}$$

Given that the wavelength $\lambda$ is $1	ext{ m}$, our path difference of $0.5	ext{ m}$ is exactly $\frac{\lambda}{2}$. In wave optics, a path difference of an odd multiple of $\frac{\lambda}{2}$ results in **destructive interference**. This perfectly aligns with the problem's statement: at point $L$, you hear a minimum intensity.

### The Journey to the Maximum

Now, the real journey begins. You start walking away from $S_1$, but you carefully maintain a constant distance of $2	ext{ m}$ from $S_2$. Geometrically, this means you are walking along a circular arc centered at $S_2$.

As you move along this arc to a new position $L'$, your distance to $S_2$ remains fixed at $2	ext{ m}$ ($S_2L' = 2	ext{ m}$). However, because you are moving away from $S_1$, your new distance to $S_1$, let's call it $d$, is increasing. Consequently, the new path difference, $\Delta x_2 = d - 2$, is also increasing.

### Finding the Distance $d$

The problem states that at this new position $L'$, you hear the **adjacent maximum** of intensity. You started at a minimum where the path difference was $\frac{\lambda}{2}$. As the path difference increases, the very next point of constructive interference (a maximum) will occur when the path difference reaches the next integer multiple of $\lambda$.

Therefore, for the adjacent maximum, the path difference must be exactly $1\lambda$:

$$\Delta x_2 = 1\lambda$$

Since $\lambda = 1	ext{ m}$, we can set up our final equation:

$$d - 2 = 1$$

Solving for $d$, we get:

$$d = 3	ext{ m}$$

And there we have it! The distance from $S_1$ where you will hear the next booming maximum is exactly $3	ext{ m}$.

Select Answer:

Visualized Solution

$S_{1}, S_{2}$ Setup

Path Difference at $L$

Calculating $S_{1} L$

Distance $S_{1} L$

Initial Path Difference

Condition for Minimum

Moving to $L^{'}$

Adjacent Maximum

Condition for Maximum

Final Answer

The Way Forward

The Sigma Insight: Interference of Waves

Visualizing the Setup

The Initial State (Minima)

The Journey to the Maximum

Finding the Distance $d$

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