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Physics - Wave on String Question with Solution | TestHub

PhysicsWave on StringWavesHard2 minQB

PhysicsHardinteger

A 50 cm wire with a mass of 4 g is stretched between supports 50.05 cm apart and vibrates at 200 Hz. The wire's cross-sectional area is 1 mm². Find $Y$ . (Answer in $P \times 1 0^{10} N / m^{2}$ ). What is the value of $P$ ?

Answer:

Solution:

$μ = 8 \times 1 0^{- 3}$ , $v = 2 \times 0.5005 \times 200 = 200.2$ m/s, $T = 320.6$ N, strain = $1 0^{- 3}$ . $Y = 320.6/ (1 0^{- 6} \times 1 0^{- 3}) = 3.2 \times 1 0^{11}$ N/m².

Stream:JEESubject:PhysicsTopic:Wave on StringSubtopic:Waves

⏱ 2mℹ️ Source: QB

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