Imagine you are in an open field where two loudspeakers are set up and connected

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.

What is the shortest distance "d" you need to walk forward to be at a point where you cannot hear the speakers?

d= ( )m

i know that da-db=1/4

but i tried to calculate by doing x^2 +3^2 = da x^2+3.5^2= db and find x and calculating from there to find the distance but it didnt work.. I only have two tries left..

Explanation / Answer

In order to solve this I'm going to assume that 1 speaker(A) is at a point (0,3) and the other (B) is at (0,-3.5) based on the calculations you were trying to do. Also, that you're walking along the x axis.

Let's first solve the distance between each speaker and the person as a function of x
DA(x)=(32+x2)

DB(x)=(3.52+x2)

Now we need to find the wavelength of the sound using the following relationship

v=f => =(v/f)=344/688

=.5m

So now we need to understand where you won't be able to hear the speakers. This will occur when the sound wave are 180 degrees out of phase. Relating this to path length means that the difference in the paths of the two sound waves differ by (2n+1)/2. Since we want to know the shortest we can make n=0.Knowing this we can set up the following equation:

/2= DB- DA

.5/2=(3.52+x2)-(32+x2)

To solve this I suggest graphing y=(3.52+x2)-(32+x2)-1/4

and finding the zeros.

x=5.625, -5.625

So the minimum distance away (from the y axis) where you hear to sound from the speakers would be 5.625m

Hope that helps

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.