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pentagon [3]
3 years ago
9

Two radio antennas A and B radiate in phase. Antenna B is a distance of 130 m to the right of antenna A. Consider point Q along

the extension of the line connecting the antennas, a horizontal distance of 50.0 m to the right of antenna B. The frequency, and hence the wavelength, of the emitted waves can be varied.(a) What is the longest wavelength for which there will be destructive interference at point ?(b) What is the longest wavelength for which there will be constructive interference at point ?
Physics
1 answer:
hoa [83]3 years ago
3 0

Answer:

constructive  λ = 130 m

destructive  λ = 260 m

Explanation:

a) the expression for constructive interference is that the difference in amino is equal to a multiple of the average wavelength

                   AQ - BQ = 2m λ/ 2       m = 0, 1, 2, ...

                     

Where

     AQ = 130 +50 = 180 m

     BQ = 50 m

To find the smallest wavelength let's use m = 1

                       180 - 50 = 2 1 λ/ 2

                        λ = 130 m

b) for destructive interference

                 AQ –BQ = (2m + 1) λ/2

The smallest value is for m = 0

                  130 = λ / 2

                  λ = 260 m

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A mass weighing 32 pounds stretches a spring 2 feet.

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(b) How many complete cycles will the mass have completed at the end of 4 seconds?

Answer:

A = 1.803 ft

Period = \frac{\pi}{2} seconds

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Explanation:

A mass weighing 32 pounds stretches a spring 2 feet;

it implies that the mass (m) = \frac{w}{g}

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= 1 slug

Also from Hooke's Law

2 k = 32

k = \frac{32}{2}

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x(0) = -1        (because of the initial position being above the equilibrium position)

x(0) = -6          ( as a result of upward velocity)

NOW, we have:

x(t)=c_1cos4t+c_2sin4t\\x^{'}(t) = 4(-c_1sin4t+c_2cos4t)

However;

x(0) = -1 means

-1 =c_1\\c_1 = -1

x(0) =-6 also implies that:

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A = \sqrt{C_1^2+C_2^2}

A = \sqrt{(-1)^2+(\frac{3}{2})^2 }

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A = 1.803 ft

Period can be calculated as follows:

= \frac{2 \pi}{4}

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3 years ago
Problem 1: Two sources emit waves that are coherent, in phase, and have wavelengths of 26.0 m. Do the waves interfere constructi
Anton [14]

1) Destructive interference

The condition for constructive interference to occur is:

\delta = m\lambda (1)

where

\delta =|d_1 -d_2| is the path difference, with

d_1 is the distance of the point from the first source

d_2 is the distance of the point from the second source

m is an integer number

\lambda is the wavelength

In this problem, we have

d_1 = 78.0 m\\d_2 = 143 m\\\lambda=26.0 m

So let's use eq.(1) to see if the resulting m is an integer

\delta =|78.0 m-143 m|=65 m\\m=\frac{\delta }{\lambda}=\frac{65 m}{26.0 m}=2.5

It is not an integer so constructive interference does not occur.

Let's now analyze the condition for destructive interference:

\delta = (m+\frac{1}{2})\lambda (2)

If we apply the same procedure to eq.(2), we find

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which is an integer: so, this point is a point of destructive interference.

2) Constructive interference

In this case we have

d_1 = 91.0 m\\d_2 =221.0 m

So the path difference is

\delta =|91.0 m-221.0 m|=130.0 m

Using the condition for constructive interference:

m=\frac{\delta }{\lambda}=\frac{130.0 m}{26.0 m}=5

Which is an integer, so this is a point of constructive interference.

3) Destructive interference

In this case we have

d_1 = 44.0 m\\d_2 =135.0 m

So the path difference is

\delta =|44.0 m-135.0 m|=91.0 m

Using the condition for constructive interference:

m=\frac{\delta }{\lambda}=\frac{91.0 m}{26.0 m}=3.5

This is not an integer, so this is not a point of constructive interference.

So let's use now the condition for destructive interference:

m=\frac{\delta}{\lambda}-\frac{1}{2}=\frac{91.0 m}{26.0 m}-0.5=3

which is an integer: so, this point is a point of destructive interference.

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