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.
Part A. 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?
a. yes
b. no
Because the path difference is equal to the wavelength of the sound, the sound originating at the two speakers will interfere constructively at your location and you will perceive a louder sound.
Part B. What is the shortest distance d you need to walk forward to be at a point where you cannot hear the speakers? The forward direction is defined as being perpendicular to a line joining the two speakers and you start walking from the line that joins the two speakers.
Express your answers in meters to three significant figures.
d = m

Answer 1

Answer:

Distance should be 5.62m

Explanation:

We know the sound to be heard louder needs constructive interference of waves and ;no sound require destructive interference !

I should be at a distance where destructive interference could take place !

a) yes of course if only one speaker is there then there will be no wave to destruct the on coming sound wave - hence it will be heard.

b) As clear from the figure if we don't want to hear sound then we should be at a point where destructive interference takes place.

The difference of distances '$r^{2}$' and '$r2^{2}$' is Δr

So, in the figure attached the Δr should be for destructive interference for no sound i-e

Δr =$\frac{(2n+1)}{2}$λ

or simply it should be like Δr=λ/2

We know formula for wavelength is  calculated by dividing speed with frequency i-e λ=v/f

and hence λ/2=v/2f

==> λ/2= 344/2×688

==>λ/2=1/4

but λ/2=Δr=$r^{2} -r2^{2}$

==> 1/4=$r^{2} -r2^{2}$

d= 5.26 m