If the distance between successive compressions and rarefaction in a sound wave is 2 m and velocity of sound is 360 m/s, then the frequency is
If the distance between successive compressions and rarefaction in a sound wave is 2 m and velocity of sound is 360 m/s, then the frequency is
Option 1 -
180 Hz
Option 2 -
45 Hz
Option 3 -
120 Hz
Option 4 -
90 Hz
Asked by Shiksha User
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1 Answer
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Correct Option - 4
Detailed Solution:Distance between successive compression and rarefaction is λ/2
.
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The acceleration of wave is g/2. Its speed increases as it moves up. So answer is (2)
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Both points B and G are moving up and are at same distance from equilibrium position at the instant.
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Kindly go through the solution 
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Equation of component waves are
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