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3 years ago

13

Two speakers face each other, and they each emit a sound of wavelength λ. One speaker is 180∘ out of phase with respect to the o

ther. If we separate the speakers by a distance 1.5λ, how far from the left-most speaker should we place a microphone in order to pick up the loudest sound? Ignore reflections from nearby surfaces. Select all that apply.. .
a) 1/4λ.
b) λ. c
) 0λ.
d) 1/2 λ. e) 3/4 λ.

1 answer:

kolbaska11 [484]3 years ago

4 0

The question ask to calculate and choose among the following is the distance of the microphone from the left - most speaker in order to pick up the loudest sound and the best answer would be letter C. I hope you are satisfied with my answer and feel free to ask for more if you have questions and further clarifications

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Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

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in this case we will call:

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G= constant of gravitation.

so:

$F_{g} =G\frac{Mm}{r^{2}}$

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

$F_{c}=ma_{c}$

where the centripetal acceleration is given by:

$a_{c}=\omega ^{2}r$

where

$\omega = \frac{2\pi}{T}$

Where T is the period, and $\omega$ is the angular speed of the planet, so:

$a_{c} = ( \frac{2\pi}{T})^{2}r$

or:

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so:

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$\sum F=ma$

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in this case we can get rid of the mass of the planet, so we get:

$G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})$

we can now solve this for $T^{2}$ so we get:

$T^{2} = \frac{4\pi ^{2}r^{3}}{GM}$

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

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So let's see what's going on there, we'll call:

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we can get rid of all the constants so we end up with:

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This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

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