Respuesta :
Answer:
- 8 m from both : Constructive
- 11 m and 7 m : Constructive
- 10 m and 8 m : Destructive
- 11 m and 14 m : In between.
- 20 m and 12 m : Constructive
- 13 m and 19 m : Destructive
- 19 m and 14 m : In between.
Explanation:
Equation of the wave
We know that the amplitude of a wave starting at [tex]\vec{x}_0[/tex] measured at position [tex]\vec{x}[/tex] at time t is
[tex]y(\vec{x},t) = A sin ( \vec{k} (\vec{x}-\vec{x}_0) - \omega t + \phi)[/tex]
where [tex]\vec{k}[/tex] is the wavevector, ω the angular frequency, and φ the phase angle.
If we measure for a time [tex]t_0[/tex] we get
[tex]y(x,t_0) = A sin ( k (x-x_0) - \omega t_0 + \phi)[/tex]
Now, we can use:
[tex]\Theta = - \omega t_0 + \phi[/tex]
[tex]y(x,t_0) = A sin ( \vec{k} (\vec{x}-\vec{x}_0) + \Theta )[/tex]
Finally, we can write this in term of the distance d, as [tex]\vec{k}[/tex] is parallel to the displacement vector for a sound wave:
[tex]\vec{k} (\vec{x}-\vec{x}_0) = k |\vec{x}-\vec{x}_0| = k d[/tex]
where k is the wavenumber
[tex]y(d,t_0) = A sin ( k d + \Theta )[/tex]
this is the amplitude of a sound wave measured at a distance d at time [tex]t_0[/tex]
Interference
Measuring two identical waves at the same time, one starting at distance d and the other at distance d', the amplitude measured is:
[tex]Amp = y(d,t_0) + y(d',t_0)[/tex]
[tex]Amp = A \ sin ( k d + \Theta ) + A \ sin ( k d' + \Theta )[/tex]
Constructive interference
We get constructive interference when both sines equals one, or minus one, so, we need a phase difference of [tex]2 n \pi[/tex], where n is an integer :
[tex]k d + \Theta = k d' + \Theta + 2 \ n \ \pi [/tex]
[tex]k d - k d'= 2 \ n \ \pi[/tex]
[tex]k (d - d')= 2 \ n \ \pi[/tex]
[tex](d - d') = \frac{2 \ n \ \pi}{k} [/tex]
as the wavenumber is
[tex]k = \frac{2\pi}{\lambda}[/tex]
where [tex]\lambda[/tex] is the wavelength,
[tex](d - d') = \frac{ \lambda \ 2 \ n \pi}{2 \pi} [/tex]
[tex](d - d') = n \lambda [/tex]
so, the difference between the distances must be a multiple of the wavelength to obtain constructive interference.
Destructive interference
We get destructive interference when one sin equals one, and the other minus one, so, we need a phase difference of [tex](2 \ n + 1) \ \pi[/tex], where n is an integer
[tex]k d + \Theta = k d' + \Theta + (2 \ n + 1) \ \pi[/tex]
[tex]k d - k d'= (2 \ n + 1) \ \pi[/tex]
[tex]k (d - d')= (2 \ n + 1) \ \pi[/tex]
[tex](d - d') = \frac{(2 \ n + 1) \ \pi}{k} [/tex]
[tex](d - d') = \frac{ \lambda (2 \ n + 1) \ \pi}{2 \pi} [/tex]
[tex](d - d') = (n + \frac{1}{2}) \lambda [/tex]
Problem
Knowing that [tex]\lambda = 4 \ m[/tex]
so, for the first
8 m from both :
[tex]d - d ' = 8 \ m - 8 \ m = 0 \ \lambda[/tex]
Constructive
11 m and 7 m:
[tex]d - d ' = 11 \ m - 7 \ m = n \ \lambda[/tex]
[tex]4 \ m = 1 * 4 \ m[/tex]
Constructive
10 m and 8 m
[tex]d - d ' = 10 \ m - 8 \ m = n \ \lambda[/tex]
[tex]2 \ m = \frac{1}{2} * 4 \ m[/tex]
Destructive
11 m and 14 m
[tex]d - d ' = 14 \ m - 11 \ m = n \ \lambda[/tex]
[tex]\frac{ d - d ' }{\lambda} = \frac{ 3 \ m }{4 \ m} = \frac{3}{4} [/tex]
In between.
20 m and 12 m
[tex]d - d ' = 20 \ m - 12 \ m = n \ \lambda[/tex]
[tex]\frac{ d - d ' }{\lambda} = \frac{ 8 \ m }{4 \ m} = 2 [/tex]
Constructive
13 m and 19 m
[tex]d - d ' = 19 \ m - 13 \ m = n \ \lambda[/tex]
[tex]\frac{ d - d ' }{\lambda} = \frac{ 6 \ m }{4 \ m} = 1 + \frac{1}{2} [/tex]
Destructive.
14 m and 19 m
[tex]d - d ' = 19 \ m - 14 \ m = n \ \lambda[/tex]
[tex]\frac{ d - d ' }{\lambda} = \frac{ 5 \ m }{4 \ m} = 1 + \frac{1}{4} [/tex]
In between.
