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bonufazy [111]
3 years ago
7

Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitt

ed by each speaker is 632 Hz . You are standing between the speakers, along the line connecting them and are at a point of constructive interference.How far must you walk toward speaker B to move to reach the first point of destructive interference? (Assume that the speed of sound in air is 344 m/s).
Physics
1 answer:
Andrews [41]3 years ago
7 0

<h2>The distance covered by man is 6.4 m</h2>

Explanation:

In the interference of two waves , these will fall upon each other .

It will make nodes and anti-nodes in between . There will be one node in between two anti-nodes and  will be one anti-node between two nodes .

The distance between two nodes is λ/2 and distance between two anti-nodes is also λ/2

Thus if the man is at constructive interference , so to move at destructive interference , it has to cover a distance equal to λ/4

Here λ = velocity/frequency = 344/632 = 18.4 m

Thus the distance he will cover = 18.4/4 = 4.6 m

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For diatomic oxygen:V=539.06 m/s

For carbon dia oxide:V=459.71 m/s

For dia atomic hydrogen:V=2156.25 m/s

Explanation:

As we know that

Root mean square velocity V

V=\sqrt{\dfrac{3RT}{M}}

Where

R is the gas constant

R=8.31\ \frac{kg.m^2}{s^2.mol.K}

T is the temperature (K).

M is the molecular weight.

For diatomic oxygen:

M=32 g/mol

T=273+100 = 373 K

R=8.31\ \frac{kg.m^2}{s^2.mol.K}

V=\sqrt{\dfrac{3RT}{M}}

V=\sqrt{\dfrac{3\times 8.31\times 373}{32\times 10^{-3}}}

V=539.06 m/s

For carbon dia oxide

M=44 g/mol

T=273+100 = 373 K

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V=\sqrt{\dfrac{3RT}{M}}

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For dia atomic hydrogen:

M= 2 g/mol

T=273+100 = 373 K

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V=\sqrt{\dfrac{3RT}{M}}

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V=2156.25 m/s

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4 years ago
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