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alexandr402 [8]
3 years ago
12

Standing waves can ruin the acoustics of a concert hall if there is excessive reflection of the sound waves that the performers

generate. For example, suppose a performer generates a 2093 Hz tone. Of a large amplitude standing wave is present, it is possible for a listener to move a distance of only 4.1 cm and hear the loudness of the tone change from loud to faint. Account for this observation in terms of standing waves, pointing out why the distance is 4.1 cm.
Physics
1 answer:
Dmitrij [34]3 years ago
6 0

Answer:

The answer to the questions is;

In terms of standing waves, the listener moves from a location with high amplitude to one with lower amplitude or vibration (anti-node to node)

The distance 4.1 cm is equivalent to λ/4

Explanation:

For standing waves we have is a stationary wave comprising of two opposite direction moving waves that have equal amplitude and frequency, resulting in the superimposition of the waves. As such certain points are fixed along the wave path that is the peaks amplitude of the wave oscillation is constant at a particular point. A node occurring at a point and an anti-node occurring at another fixed point

When the listener moves 4.1 cm he or she has left the anti-node to the node hence the faintness of the sound

The distance from the node to the anti-node is 1/4 wavelength, or 1/4×λ

Therefore 4.1 cm is λ/4

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<u>Given the following data;</u>

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Mathematically, speed is given by the formula;

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Making distance the subject of formula, we have;

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Substituting into the above formula, we have;

Distance = 25 * 2.8

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A loop of wire in the shape of a rectangle rotates with a frequency of 284 rotation per minute in an applied magnetic field of m
yaroslaw [1]

Answer:

a) Magnitude of maximum emf induced = 0.0714 V = 71.4 mv

b) Maximum current through the bulb = 0.00793 A = 7.93 mA

Explanation:

a) The induced emf from Faraday's law of electromagnetic induction is related to angular velocity through

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The maximum emf occurs when (sin wt) = 1

Maximum Emf = NABw

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w = (284/60) × 2π = 29.75 rad/s

E(max) = 1×0.0004×6×29.75 = 0.0714 V = 71.4 mV

Note that: since we're after only the magnitude of the induced emf, the minus sign that indicates that the induced emf is 8n the direction opposite to the change in magnetic flux, is ignored for this question.

b) Maximum current through the bulb

E(max) = I(max) × R

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E(max) = 0.0714 V

I(max) = ?

0.0714 = I(max) × 9

I(max) = (0.0714/9) = 0.00793 A = 7.93 mA

Hope this Helps!!

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