Sound source A is located at x = 0, y = 0, and sound source B is located at x = 0, y = 2.3 m. The two sources radiate coherently
and in phase. An observer at x = 17.8 m, y = 0 notes that as he takes a few steps from y = 0 in either the +y or −y direction, the sound intensity increases. What is then the lowest frequency and next to lowest frequency of the sources that can account for this observation? (Enter your answers from smallest to largest.)
1 answer:
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Answer:
1) tensile stress = 76.648 Mpa
2) extension = 0.0215 m
Explanation:
Detailed explanation and calculation is shown in the image below
I believe this is what you have to do:
The force between a mass M and a point mass m is represented by
So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁
So F₁ = G(Mm/r^2)
Now the distance has doubled so lets account for this in F₂:
F₂ = G(Mm/(2r)^2)
Now square the 2 that gives you four and we can pull that out in front to give
F₂ = G(Mm/r^2)
Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations
now we see that:
F₂ = F₁
So the second force will be 0.25 (1/4) x 1600 or 400 N.