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Physics - Wave on String Question with Solution | TestHub

PhysicsWave on StringWavesHard2 minQB

PhysicsHardsingle choice

A rubber string of natural length 20 cm obeying Hooke\'s law is stretched to 24 cm, giving lowest resonance frequency $n_{0}$ . It is then allowed to contract to 22 cm. The lowest frequency now is:

Options:

Answer:

Solution:

$n \propto (L - 20) / L$ . At 24 cm: $4/24 = 0.167$ . At 22 cm: $2/22 = 0.091$ , which is lower, so $n < n_{0}$ .

Stream:JEESubject:PhysicsTopic:Wave on StringSubtopic:Waves

⏱ 2mℹ️ Source: QB

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