Two loudspeakers are 2.54 m apart. A person stands 3.04 m from one speaker and 3.50 m from the other. What is the lowest frequency at which destructive interference (sound cancellation) will occur at this point?
distances have to be one half wavelength apart.
3.04-3.50=1/2 lambda
lambda= .92meter check that, so
f= velocitysound/.92
To find the lowest frequency at which destructive interference will occur at the given point, we need to determine the condition for destructive interference. Destructive interference happens when the path difference between the two sources is equal to an odd multiple of half of the wavelength.
Let's denote:
- Distance between the two loudspeakers as d (d = 2.54 m)
- Distance from the first speaker to the person as x (x = 3.04 m)
- Distance from the second speaker to the person as y (y = 3.50 m)
- Velocity of sound in air as v
We can begin by calculating the path difference between the two speakers at the given point. The path difference (Δx) can be calculated using the formula:
Δx = y - x
Now, we need to determine the distance traveled by the sound wave during one wavelength. This is given by:
Length of one wavelength = v/frequency
Since we are looking for the lowest frequency, we'll assume that the wavelength is at its maximum, which is twice the distance between the speakers (2d). So:
Length of one wavelength = 2d = 2 * 2.54 = 5.08 m
Now, we can equate the path difference to an odd multiple of half of the wavelength and solve for the frequency:
Δx = λ/2
Substituting the values:
y - x = (5.08/2) * (2n + 1)
Now, we can rearrange the equation to solve for the frequency (f):
f = v / λ
Substituting the value of λ from earlier and rearranging, the equation becomes:
f = v / (2d * (2n + 1))
Finally, we can substitute the given value of v (velocity of sound in air) which is approximately 343 m/s:
f = 343 / (2 * 2.54 * (2n + 1))
To find the lowest frequency, we need to find the smallest value of n that satisfies the condition for destructive interference. By plugging in different values of n, we can determine the lowest frequency.