Question

Two speakers that are 12.0 m apart produce in-phase sound waves of frequency 245 Hz in a room where the speed of sound is 340 m/s. A woman starts out at the midpoint between the two speakers. The room's walls and ceiling are covered with absorbers to eliminate reflections, and she listens with only one ear for best precision.(a) Why is this constructive interference?(b) She now walks slowly toward one of the speakers. How far from the center must she walk before she first hears the sound reach a minimum intensity?(c) How far from the center must she walk before she first hears the sound maximally enhanced?

Answer 1

Answer:

a) Please see below as the answer is self-explanatory

b) 0.35 m

c) 0.7 m

Explanation:

a) As there is no difference in the path (because she is exactly at the midpoint between both speakers) that the waves do in order to reach to her ear, she listens both waves with the same intensity.

The same effect would be experimented, if the difference between paths, were an even multiple of the half of the wavelength.

b) In order to have a destructive interference, the difference in path (in the worst case) must be equal to an odd multiple of the half wavelength.

In any wave, there is a fixed relationship between speed, frequency and wavelength, as follows:

λ = v / f

Replacing with our givens for frequency (245 Hz) and v (340 m/s), we have:

λ = 340 m/s / 245 Hz = 1.4 m

The condition for destructive interference, is as follows:

Δd = d₂-d₁ = (2n + 1) *λ/2

For n=0, we have the shortest distance:

Δd = λ/2 = 0.7 m

So, in order to have such a path difference, she must walk exactly the half of this distance from the center, i.e., 0.35 m.

c) With the same reasoning as above, as the condition for constructive interference, as we have already mentioned, is that the path difference be at least one full wavelength, we can say:

Δd = d₂-d₁ = λ

So, in order to have such a path difference, she must walk exactly the half of this distance from the center, i.e., 0.7 m.

Answer 2

In this exercise we have to use the knowledge of waves and their interference to calculate the values ​​that occur:

a) The same effect would be experimented, if the difference between paths, were an even multiple of the half of the wavelength.

b) 0.35 m

c) 0.7 m

What is wave interference?

Wave interference is the superposition of two waves in space. This phenomenon can be classified in two ways: constructive or destructive interference.

organizing the information given in the statement we have:

• 12.0 m apart produce
• frequency 245 Hz
• speed of sound is 340 m/s

a) Destructive interference – occurs when waves do not have the same phase and have an annihilation character.

B) then remembering the destructive wave formula we have that:

$\lambda = v / f$

Putting the given values ​​in the statement we have:

$\lambda = 340 / 245 = 1.4 m$

The condition for destructive interference, is as follows:

$\Delta d = d_2-d_1 = (2n + 1) *\lambda/2\\\Delta d = 0.35 m$

c) The condition for destructive interference, is as follows:

$\Delta d = d_2-d_1 = (2n + 1) *\lambda/2\\\Delta d = 0.7 m$

See more about destructive interference at brainly.com/question/16098226