# Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the...

## Question:

Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are {eq}32\ cm {/eq} apart. The sound intensity decreases as the distance between the speakers is increased, reaching zero at a separation of {eq}40\ cm {/eq}.

A) What is the wavelength of the sound?

B) If the distance between the speakers continues to increase, at what separation will the sound intensity again be a maximum?

## Interference of Waves:

When two waves are traveling in the same medium, the two waves combine to give out a resultant wave (if they are of the same nature). This phenomenon is called the interference of waves. It is of two types:

• Constructive Interference- This happens when the crest of one wave falls over the crest of the other wave. The amplitude of the resultant wave is the sum of the amplitudes of the original waves.
• Destructive Interference- This happens when the crest of one wave falls over the trough of the other wave. The amplitudes, in this case, subtract. The amplitude of the resultant wave is zero if the waves are identical.

If the 2 waves are identical, then the type of interference would depend only on the relative distance of the point of inspection from the two sources.

When two waves of same frequency/wavelength travel along a line, the intensity of the resultant wave is maximum at a point if the difference in the path traveled by the waves to reach the point is an integral multiple of their wavelength i.e

{eq}\Delta x=n\lambda\;\;\;\;\;\;\;\;\text{(Condition for Maxima)} {/eq}

Here,

• {eq}\Delta x {/eq} is the path difference between the two waves.
• {eq}n=0,1,2,3,4.. {/eq} is an integer.
• {eq}\lambda {/eq} is the wavelength of the wave.

For minimum intensity, the path difference must be:

{eq}\Delta x=\left ( n+\dfrac{1}{2} \right )\lambda\;\;\;\;\;\;\;\;\text{(Condition for Minima)} {/eq}

According to the question:

{eq}32=n\lambda\;\;\;\;\;\;\;\;\text{(1)}\\ 40=\left ( n+\dfrac{1}{2} \right )\lambda\;\;\;\;\;\;\;\;\text{(2)} {/eq}

On subtracting (1) from (2)

{eq}\begin{align*} 40-32&=n\lambda+\dfrac{\lambda}{2}-n\lambda\\ \Rightarrow \dfrac{\lambda}{2}&=8\\ \Rightarrow \lambda&=16\;\rm cm \end{align*} {/eq}

b)

The next nearest separation distance for maxima would correspond to the next integer after {eq}n {/eq}, (say {eq}m {/eq}). Then

{eq}\begin{align*} \Delta x&=m\lambda\\ &=\left ( n+1 \right )\lambda\\ &=n\lambda+\lambda\\ &=32+\lambda\\ &=32+16\\ &=48\;\rm cm \end{align*} {/eq}