Question

Imagine you are in an open field where two loudspeakers are set up and connected to the same amplifier so that they emit sound waves in phase at 688 Hz. Take the speed of sound in air to be 344 m/s.If you are 3.00 m from speaker A directly to your right and 3.50 m from speaker B directly to your left, will the sound that you hear be louder than the sound you would hear if only one speaker were in use?
What is the shortest distance d you need to walk forward to be at a point where you cannot hear the speakers?

Answer 1

Answer:

yes it will be louder sound than he use only one speaker

so he must have to move towards left speaker by distance 0.125 m

Explanation:

As we know that the two speakers are coherent source of sound

So here we can say that the sound received from the two sources will give maximum intensity when path difference of the source is integral multiple of the wavelength of the sound

While if we receive minimum intensity of sound then the path difference from each source must be odd multiple of half the wavelength of sound.

so here we know that

$wavelength = \frac{speed}{frequency}$

$\lambda = \frac{344}{688}$

$\lambda = 0.5 m$

Now path difference from each speaker is given as

$\Delta x = L_1 - L_2$

$\Delta x = 3.50 - 3.00$

$\Delta x = 0.50 m$

So this is integral multiple of wavelength and hence it will be the position of maximum intensity of sound

So yes it will be louder sound than he use only one speaker

Now if he wish to be at the position of No sound

then we have

$\Delta x = \frac{\lambda}{2}$

$\Delta x = \frac{0.50}{2} = 0.25$

so he must have to move towards left speaker by distance 0.125 m