Question

Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 5.00 m. A person is standing 12.0 m away from the wall, equidistant from the loudspeakers. When the person moves 50 cm parallel to the wall, she experiences destructive interference for the first time.

What is the frequency of the sound?

Answer 1

Answer: Frequency = 42Hz

Explanation: first destructive will be:

(1 + 0.5)× V/ lamda = F

Please find the attached file for the solution

Answer 2

841.5 Hz

Step-by-step explanation:

Given

y = 50 cm = 0.5 m

d = 5.00 m

L = 12.0 m away from the wall

v = speed of sound = 343 m/s

The image of the scenario is presented in the attached image.

When destructive interference is being experienced from 50 cm (0.5 m) parallel to the wall, the path difference between the distance of the two speakers from the observer is equal to half of the wavelength of the wave.

Let the distance from speaker one to the observer's new position be d₁

And the distance from the speaker two to the observer's new position be d₂

(λ/2) = |d₁ - d₂|

d₁ = √(12² + 3²) = 12.3693 m

d₂ = √(12² + 2²) = 12.1655 m

|d₁ - d₂| = 0.2038 m

(λ/2) = |d₁ - d₂| = 0.2038

λ = 0.4076 m

For waves, the velocity (v), frequency (f) and wavelength (λ) are related thus

v = fλ

f = (v/λ) = (343/0.4076) = 841.5 Hz

Hope this Helps!!!