Question

Two speakers (labelled 1 and 2) emit a sound at a wavelength of λ=4 m. They produce destructive interference at point P, which is 3 m from speaker 1 and 1 m from speaker 2 . What can you say about the phase constant difference of the two speakers? They have an initial phase phase difference Δϕ0​=23π​ They are completely in phase (Δϕ0​=0) They have an initial phase phase difference Δϕ0​=2π​ More than one of these answers is possible. They are completely out of phase (Δϕ0​=π) A musician is attempting to tune his guitar using a tuning fork. He hits the tuning fork which produces a sound of 440 Hz. He then plays a note on his guitar which produces a sound of 436 Hz. How many beats (loud-quiet-loud cycles) will he hear in 10 seconds? 2.5 0.4 40 /canvas.ubc.ca/courses/104277/quizzes/524849 12/623:24 Quiz 5: PHYS 131 ALL SECTIONS 2022W1 Energy and Waves (

Answer 1

The speakers have an initial phase difference of π radians (180°), meaning they are completely out of phase, resulting in destructive interference at point P. The musician will hear 40 beats over the span of 10 seconds from the guitar and tuning fork.

Step-by-step explanation:

For two speakers to produce destructive interference at a point P, the difference in the path lengths that the sound waves travel from each speaker to point P must be an odd multiple of half a wavelength, or (2n + 1)λ/2, where n is an integer. Given that the wavelength λ = 4 m and speaker 1 is 3 m from point P while speaker 2 is 1 m from point P, the path difference is 3 m - 1 m = 2 m, which is exactly λ/2. This means that the waves from the two speakers arrive at point P with a phase difference of π rad (180°), leading to destructive interference, implying they are completely out of phase (∆φ0​=π).

For the beat frequency created by two sounds of 440 Hz and 436 Hz, one can calculate it as the difference between the two frequencies, which is |440 Hz - 436 Hz| = 4 Hz. Therefore, over 10 seconds, the musician will hear 4 beats per second × 10 seconds = 40 beats.

Answer 2

The speakers likely have an initial phase difference of π radians for destructive interference at point P. The musician will hear 40 beats in 10 seconds from the guitar and the tuning fork.

Step-by-step explanation:

For two speakers emitting sound waves that produce destructive interference at point P, the phase difference between the speakers must be an odd multiple of π radians (180°). Given the wavelength λ=4 m, and the distances from point P to speakers 1 and 2 being 3 m and 1 m respectively, the path difference is 2 m, which is half a wavelength (λ/2).

This leads to destructive interference. Considering the initial phase difference, it is most likely that the speakers have an initial phase difference Δφ₀ = π because this represents an out-of-phase condition that would cause destructive interference at point P.

For the wave beats question, when two sound waves of frequencies 440 Hz and 436 Hz are played together, the number of beat frequencies heard in one second is the absolute difference in their frequencies, which is 4 beats per second. Therefore, over 10 seconds, he will hear 40 beats.