Two small stereo speakers A and B that are 1.4 m apart are sending out sound of wavelength 34 cm in all directions and all in phase. A person at point P starts out equidistant from both speakers and walks so that he is always 1.5 m from speaker B. For what values x will the sound this person hears be (a) Maximally reinforced, (b) cancalled? Limit your solution to the cases where x is less then or equal to 1.5 m.
Can someone please help, I am totally confused!
this question just sent me into an existential crisis
To find the values of x for which the sound heard by the person at point P is maximally reinforced or cancelled, we can use the concept of constructive and destructive interference.
Let's start by understanding the scenario. We have two speakers, A and B, separated by a distance of 1.4 m. They emit sound waves of a wavelength of 34 cm (or 0.34 m) in all directions and are in phase, meaning they start at the same point in the wave cycle.
The person at point P initially stands equidistant from both speakers, which means their distance from each speaker is the same (let's call it d). As the person walks, they maintain a constant distance of 1.5 m from speaker B.
(a) Maximally Reinforced Sound:
To find the values of x for which the sound is maximally reinforced, we need to consider the condition for constructive interference.
Constructive interference occurs when the path difference, represented by Δx, between the two speakers is equal to an integral multiple of the wavelength (mλ). In other words, Δx = mλ, where m is an integer.
Initially, the person is equidistant from both speakers (d = 1.5 m). As they move, the distance between them and speaker A becomes (d + x), where x is the distance the person has moved away from the midpoint.
So, the path difference Δx is given by:
Δx = (d + x) - d = x
To find the values of x for maximal reinforcement, we can set:
x = mλ, where m is an integer.
Given that x must be less than or equal to 1.5 m, we have:
x ≤ 1.5 m
Substituting the wavelength (λ = 0.34 m), we get:
mλ ≤ 1.5 m
m(0.34 m) ≤ 1.5 m
m ≤ 4.41
Since m must be an integer, the possible values for m in this case are 1, 2, 3, and 4. Therefore, the values of x for which the sound heard by the person at point P is maximally reinforced are 0.34 m, 0.68 m, 1.02 m, and 1.36 m.
(b) Cancelled Sound:
To find the values of x for which the sound is cancelled, we need to consider the condition for destructive interference.
Destructive interference occurs when the path difference, represented by Δx, between the two speakers is equal to an odd multiple of half the wavelength [(2m + 1) * (λ/2)], where m is an integer.
Using the same approach as above, we find:
x = (2m + 1) * (λ/2)
Again, since x must be less than or equal to 1.5 m, we have:
x ≤ 1.5 m
Substituting the wavelength (λ = 0.34 m), we get:
(2m + 1) * (0.34/2) ≤ 1.5 m
(2m + 1) * 0.17 ≤ 1.5 m
(2m + 1) ≤ 8.82
Simplifying the inequality, we have:
2m + 1 ≤ 8.82
2m ≤ 7.82
m ≤ 3.91
Since m must be an integer, the possible values for m in this case are 1, 2, and 3. Therefore, the values of x for which the sound heard by the person at point P is cancelled are approximately 0.17 m, 0.68 m, and 1.19 m.
To summarize:
(a) The sound will be maximally reinforced for x values of 0.34 m, 0.68 m, 1.02 m, and 1.36 m.
(b) The sound will be cancelled for x values of approximately 0.17 m, 0.68 m, and 1.19 m.