HardPhysicsHigh School (11–12)Waves & Sound › Superposition and Standing WavesNumericalJEE Advanced JEE Main

Two small loudspeakers S₁ and S₂ stand 1.5 m apart in an open field and are fed by the same…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Two small loudspeakers $S_1$ and $S_2$ stand $1.5\ \text{m}$ apart in an open field and are fed by the same oscillator, so they send out sound of the same frequency, in phase, with the same power. Each spreads its sound equally in all directions. A listener stands at $P$ , $2.0\ \text{m}$ from $S_1$ , on the line through $S_1$ at right angles to $S_1S_2$ , as in the figure. The speed of sound is $340\ \text{m/s}$ and the frequency is $1020\ \text{Hz}$ .

Top view of a field. Two loudspeakers S1 and S2 lie on a horizontal line, 1.5 m apart, with S1 on the left. A listener P stands 2.0 m from S1, straight above S1 in the drawing, so the line S1 P is at right angles to S1 S2. A small square marks the right angle at S1. Dashed lines join P to each speaker.

What is the path difference at $P$ , in wavelengths?
What is the intensity at $P$ , as a fraction of the intensity that $S_1$ alone would give there?
The frequency is raised slowly. At what frequency is the sound at $P$ next loudest, the two waves arriving in phase?
What is the lowest frequency at which the two waves arrive at $P$ in opposite phase?
Back at $1020\ \text{Hz}$ , $S_2$ is rewired so that it runs a quarter of a period ahead of $S_1$ . What is the intensity at $P$ now, as a fraction of the intensity of $S_1$ alone?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Find the two distances, the phase difference they give, and the two intensities, then add the waves with $I = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\delta$ .

Step 1. Path difference (a). $PS_2 = \sqrt{(1.5\ \text{m})^2 + (2.0\ \text{m})^2} = 2.5\ \text{m}$ , so $\Delta = 0.5\ \text{m}$ . With $\lambda = 340\ \text{m/s} / 1020\ \text{Hz} = 33.3\ \text{cm}$ , $\Delta/\lambda = 1.5$ .

Step 2. Intensity (b). $\dfrac{I_2}{I_1} = \left(\dfrac{2.0\ \text{m}}{2.5\ \text{m}}\right)^2 = 0.64$ . The phase difference is $2\pi \times 1.5 = 3\pi$ , so $\cos\delta = -1$ and

\displaystyle \frac{I}{I_1} = 1 + 0.64 - 2\sqrt{0.64} = 0.04

Nearly silent, but not quite: the amplitudes differ.

Step 3. Next loud note (c). The waves arrive in phase when $\Delta$ is a whole number of wavelengths: $f = n\dfrac{c}{\Delta} = n \times 680\ \text{Hz}$ . Above $1020\ \text{Hz}$ the next is $1360\ \text{Hz}$ .

Step 4. Lowest opposite phase (d). Opposite phase needs $\Delta = \lambda/2$ at least, so $f = \dfrac{c}{2\Delta} = 340\ \text{Hz}$ .

Step 5. A quarter-period lead (e). The lead changes the phase difference by $\pi/2$ , so $\delta = 3\pi \pm \pi/2$ and $\cos\delta = 0$ . Then $I/I_1 = 1 + 0.64 = 1.64$ : the intensities simply add.

Answer: (a) $1.5$ . (b) $0.04$ . (c) $1360\ \text{Hz}$ . (d) $340\ \text{Hz}$ . (e) $1.64$ .

Watch out. With equal amplitudes, (b) would give zero, and a loud point would give $4I_1$ . Here a loud point gives $(1 + 0.8)^2 I_1 = 3.24\,I_1$ .

← All Problems

← A sonometer wire is 0.80 m long between its bridges, and its tension comes from a load hanging over…Next →

More onWaves & Sound › Superposition and Standing Waves

Discussion (0)

Loading…