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Physics - Wave on String Question with Solution | TestHub

PhysicsWave on StringWavesHard2 minQB

PhysicsHardsingle choice

A 50 cm wire of mass 4 g and cross section 1 mm² is stretched between supports 50.1 cm apart. Its fundamental frequency is 25 Hz. Find $Y$ in the form $X \times 1 0^{9}$ N/m² .

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Answer:

Solution:

$μ = 4 \times 1 0^{- 3} /0.5 = 8 \times 1 0^{- 3}$ kg/m. $v = 2 \times 0.501 \times 25 = 25.05$ m/s. $T = 8 \times 1 0^{- 3} \times 627.5 = 5.02$ N. Strain = $0.1/50 = 2 \times 1 0^{- 3}$ . $Y = 5.02/ (1 0^{- 6} \times 2 \times 1 0^{- 3}) = 2.51 \times 1 0^{9}$ N/m², so $X \approx 2.5$ .

Stream:JEESubject:PhysicsTopic:Wave on StringSubtopic:Waves

⏱ 2mℹ️ Source: QB

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