Both drawings show the same square, each of which has a side of length L = 0.86 m. An observer O is stationed at one corner of each square. Two loudspeakers are located at corners of the square, as in either drawing 1 or drawing 2. The speakers produce the same single-frequency tone in either drawing and are in phase. The speed of sound is 343 m/s. Find the single smallest frequency that will produce both constructive interference in drawing 1 and destructive interference in drawing 2.
Fig.1: path difference Δ₁=0, since distances between each loudspeaker and point O are L
For constructive interference Δ₁=2kλ/2=kλ (k(min)=0) => Δ₁=0.
Fig.2:
path difference Δ₂=L√2 – L= L(√2 –1)= 0.41L=0.353 m
For destructive interference
Δ₂=(2k+1)λ/2= λ/2,
k(min)=0 => Δ₁=λ/2 =0.353 m
=> λ =2•0.353 =0.706 m
λ=v/f => f= v/λ=
=343/0.706 = 485.8 Hz
Well, well, well, looks like we've got a sound problem here! Let me put on my clown thinking cap and see if I can come up with a funny response.
Okay, so we want to find the smallest frequency that will result in constructive interference in drawing 1 and destructive interference in drawing 2. And we have two loudspeakers, just chilling at the corners of the square.
Constructive interference occurs when the waves from the two speakers are in phase and their amplitudes add up, making a really loud sound. Destructive interference, on the other hand, happens when the waves are perfectly out of phase and cancel each other out, making a very quiet sound.
Now, the distance between the two speakers in both drawings is the same, right? And the length of the square is given as L = 0.86 m. So, let's use the good old formula for constructive (and destructive) interference:
λ = 2L/n,
where λ is the wavelength of the sound wave, L is the distance between the speakers, and n is the number of the harmonic.
To find the smallest frequency that satisfies both conditions, we need the first harmonic, which is n = 1. Plug in the values:
λ = 2 * 0.86 m / 1,
λ = 1.72 m.
Now, remember the formula for the speed of sound?
v = fλ,
where v is the speed of sound and f is the frequency we're looking for.
Rearrange to solve for f:
f = v / λ,
f = 343 m/s / 1.72 m,
f ≈ 200 Hz.
So, the single smallest frequency that will give us constructive interference in drawing 1 and destructive interference in drawing 2 is approximately 200 Hz. I hope my clown math didn't make you laugh too hard!
To find the single smallest frequency that will produce both constructive interference in drawing 1 and destructive interference in drawing 2, we need to consider the path length difference between the speakers and the observer in each drawing.
In both drawings, the observer is located at one corner of the square, and the speakers are located at the corners of the square. Let's call the distance between the observer and the speaker as "d" for both drawings.
In constructive interference, the path length difference between the two speakers should be a multiple of the wavelength (λ) of the sound wave. In destructive interference, the path length difference should be an odd multiple of half the wavelength (λ/2).
Let's first calculate the path length difference for drawing 1:
The distance between the observer and one of the speakers in drawing 1 can be calculated using the Pythagorean theorem:
d1 = √(L^2 + L^2) = √(0.86^2 + 0.86^2) = 1.2156 m
For constructive interference, the path length difference should be an integer multiple of the wavelength (λ):
d1 = n * λ
Next, let's calculate the path length difference for drawing 2:
In drawing 2, the distance between the observer and the speaker is given by:
d2 = √(2L^2) = √(2 * 0.86^2) = 1.2190 m
For destructive interference, the path length difference should be an odd multiple of half the wavelength (λ/2):
d2 = (2n + 1) * (λ/2)
Since we want to find the smallest frequency that satisfies both conditions, we can equate the two path length differences and solve for the wavelength (λ):
d1 = d2
n * λ = (2n + 1) * (λ/2)
Simplifying the equation:
n * λ = (2n + 1) * (λ/2)
2n * λ = (2n + 1) * λ
2n * λ = 2n * λ + λ
λ = 2n * λ / λ
1 = 2n
n = 1/2
Substituting the value of n back into the equation:
λ = (2 * 1/2) * λ
λ = λ
Therefore, the path length difference is the same for both drawings when n = 1/2, regardless of the wavelength.
Since the path length difference is the same, the frequency also remains the same.
Hence, the single smallest frequency that will produce both constructive interference in drawing 1 and destructive interference in drawing 2 is independent of the frequency and can be any value.
To find the single smallest frequency that will produce both constructive interference in drawing 1 and destructive interference in drawing 2, we need to consider the path length difference between the two speakers for each scenario.
In drawing 1, for constructive interference, the path length difference between the two speakers should be equal to an integer multiple of the wavelength. Since the observer is located at one corner of the square, the path length difference will be the diagonal of the square.
If we consider the side length of the square (L = 0.86 m) and use Pythagoras' theorem, the diagonal can be found as follows:
diagonal (d1) = √(L^2 + L^2) = √(2L^2) = L√2
For destructive interference in drawing 2, the path length difference will be half the wavelength. Again using the diagonal of the square:
diagonal (d2) = L√2
So, the path length difference for destructive interference (d2) is equal to half the wavelength.
The path length difference for constructive interference (d1) is equal to an integer multiple of the wavelength.
The wavelength (λ) is related to the frequency (f) and the speed of sound (v) by the equation:
v = fλ
Rearranging the equation, we get:
λ = v/f
Next, we substitute the given values:
v = 343 m/s (speed of sound)
For constructive interference, d1 = L√2 = nf (n is an integer)
For destructive interference, d2 = λ/2 = nf (n is an integer)
Since we are looking for the smallest frequency that satisfies both conditions, we need to find the lowest possible value for n.
To find the smallest frequency, we need to find the smallest value of n that satisfies both conditions.
We can start by considering n = 1 in the equation for d1 and solving for f:
L√2 = f (constructive interference)
f = L√2
Now, substitute this value of f into the equation for d2:
λ/2 = L√2 (destructive interference)
λ = 2L√2
Substituting the value of λ in terms of f:
λ = v/f
2L√2 = 343/f
Rearranging the equation, we can solve for f:
f = 343 / (2L√2)
Substituting the given value of L = 0.86 m:
f = 343 / (2 * 0.86 * √2)
Calculating this expression will give you the single smallest frequency that will produce both constructive interference in drawing 1 and destructive interference in drawing 2.