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Question

Stationary waves of frequency 200Hz are formed in the air. If the velocity of the wave is 360m/s, the shortest distance between two antinodes is


A

1.8m

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B

3.6m

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C

0.9m

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D

0.45m

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Solution

The correct option is C

0.9m


Step 1: Given data

The shortest distance between two antinodes is equal to the half of wavelength.

Here v=360m/s and f=200Hz

Step 2: Formula used

Let λ=wavelength

λ=vf

Where f is the frequency and v is the velocity of the wave

Step 3: Calculate the shortest distance between the antinodes

λ=vf=360200=1.8m

So λ2=1.82=0.9m

Hence, option C is correct.


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