A wire of density 9 × 10 – 3 kg cm – 3 is stretched between two clamps 1 m apart. The resulting strain in the wire is 4 . 9 × 10 – 4 . The lowest frequency of the transverse vibrations in the wire (Young's modulus of wire Y = 9 × 10 10 Nm – 2 ), (to the nearest integer),_______

Fundamental frequency in the string f = 1 2 ℓ T μ = 1 2 ℓ T ρA = 1 2 ℓ YΔℓ ρℓ f = 1 2 ℓ YΔℓ ρℓ = 1 2 9 × 10 10 × 4 . 9 × 10 - 4 9 × 10 - 3 × 10 - 6 = 35

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Physics - Wave on String Question with Solution | TestHub

PhysicsWave on StringStandingMedium2 minPYQ_2020

PhysicsMediumnumerical

A wire of density $9 \times 10^{- 3} kg {cm}^{- 3}$ is stretched between two clamps $1 m$ apart. The resulting strain in the wire is $4.9 \times 10^{- 4}$ . The lowest frequency of the transverse vibrations in the wire (Young's modulus of wire $Y = 9 \times 10^{10} {Nm}^{- 2}$ ), (to the nearest integer),_______

Answer:

35.00

Solution:

Fundamental frequency in the string

$f = \frac{1}{2 ℓ} \sqrt{\frac{T}{μ}} = \frac{1}{2 ℓ} \sqrt{\frac{T}{ρA}} = \frac{1}{2 ℓ} \frac{\sqrt{YΔℓ}}{ρℓ}$
$f = \frac{1}{2 ℓ} \sqrt{\frac{YΔℓ}{ρℓ}}$
$= \frac{1}{2} \sqrt{\frac{9 \times 10^{10} \times 4.9 \times 10^{- 4}}{9 \times 10^{- 3} \times 10^{- 6}}} = 35$

Stream:JEESubject:PhysicsTopic:Wave on StringSubtopic:Standing

⏱ 2mℹ️ Source: PYQ_2020

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