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denpristay [2]
3 years ago
14

Sound with a frequency of 1250 Hz leaves a room through a doorway with a width of 1.05 m.At what minimum angle relative to the c

enterline perpendicular to the doorway will someone outside the room hear no sound? Use 344 m/s for the speed of sound in air and assume that the source and listener are both far enough from the doorway for Fraunhofer diffraction to apply. You can ignore effects of reflections.Theta = ? degrees, ? degrees, ? degrees (should be three terms)
Physics
1 answer:
kiruha [24]3 years ago
5 0

Answer:

15.19°, 31.61°, 51.84°

Explanation:

We need to fin the angle for m=1,2,3

We know that the expression for wavelenght is,

\lambda = \frac{c}{f}

Substituting,

\lambda = \frac{344}{1250}

\lambda = 0.2752m

Once we have the wavelenght we can find the angle by the equation of the single slit difraction,

sin\theta = \frac{m \lambda}{W}

Where,

W is the width

m is the integer

\lambda the wavelenght

Re-arrange the expression,

\theta = sin^{-1} \frac{m\lambda}{W}

For m=1,

\theta = sin^{-1} \frac{1 (0.2752)}{1.05}= 15.19\°

For m=2,

\theta = sin^{-1} \frac{2 (0.2752)}{1.05}= 31.61\°

For m=3,

\theta = sin^{-1} \frac{3 (0.2752)}{1.05}= 51.84\°

<em>The angle of diffraction is directly proportional to the size of the wavelength.</em>

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Given the data in the question

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To find the Tension in each of the ropes, we make use of the equation from Newton's Second Laws of Motion:

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So Tension of the rope;

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Therefore, the tension in the first rope is 147 Newton

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Since only the block of mass 10kg is hang from the second, the tension in the second rope will be;

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Learn More, brainly.com/question/18288215

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Answer:

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6 0
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