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Whitepunk [10]
3 years ago
10

A microphone is located on the line connecting two speakers that are 0.513 m apart and oscillating in phase. The microphone is 1

.80 m from the midpoint of the two speakers. What are the lowest two frequencies that produce an interference maximum at the microphone's location?
Physics
1 answer:
pantera1 [17]3 years ago
7 0

Answer:

frequency 1 = 334.30 Hz

frequency 2 = 1002.92 Hz

Explanation:

Given data

speaker distance y = 0.513 m

microphone distance D = 1.80 m

to find out

lowest two frequencies

solution

we know velocity of sound is 343 m/s

so we consider point x

so at 1st speaker distance from x   = D + (y/2)

1st speaker distance from x   = 1.80 + (0.513/2) = 2.0565 m   .....1

and

at 2nd speaker distance from x   = D - (y/2)

2nd speaker distance from x   = 1.80 - (0.513/2) = 1.5435 m     .........2

so destructive interference from 1 and 2  we know

1st - 2nd = ( m + 0.5 ) wavelength

2.0565 m - 1.5435 m = (  0+ 0.5) wavelength

wavelength  = 1.026 m

so here 1st min frequency will be

frequency 1 = velocity of sound / wavelength

frequency 1 = 343 / 1.026 =334.30 Hz

and

2nd  min frequency will be

frequency 2 =

2.0565 m - 1.5435 m = (  1 + 0.5) wavelength

wavelength  = 0.342 m

frequency 2 =  velocity of sound / wavelength

frequency 2 = 343 / 0.342 = 1002.92 Hz

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Answer:

See the answers below.

Explanation:

In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].

E_{A}=E_{B}

The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.

So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.

E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}

At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.

E_{B}=m*g*h+\frac{1}{2}*m*v^{2}

Therefore we will have the following equation:

(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]

The kinetic energy can be easily calculated by means of the kinetic energy equation.

KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]

In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.

E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]

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