A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 × 10 3 kg m - 3 and its Young's modulus is 9.27 × 10 10 Pa . What will be the fundamental frequency of the longitudinal vibrations?

L = ⋋ 2 ⇒ ⋋ = 2 L = 1.2 m v = y ρ = 5.86 × 10 3 m s - 1 ∴ Fundamental frequency = v ⋋ = 4.88 kHz ≈ 5 kHz

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PhysicsWave on StringEquation/graphMedium2 minPYQ_2018

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A granite rod of $60 cm$ length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is $2.7 \times 10^{3} kg m^{- 3}$ and its Young's modulus is $9.27 \times 10^{10} Pa$ . What will be the fundamental frequency of the longitudinal vibrations?

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Answer:

Solution:

$L = \frac{⋋}{2} \Rightarrow ⋋ = 2 L = 1.2 m$

$v = \sqrt{\frac{y}{ρ}} = 5.86 \times 10^{3} m s^{- 1}$

$∴ Fundamental frequency = \frac{v}{⋋} = 4.88 kHz$

$\approx 5 kHz$

Stream:JEESubject:PhysicsTopic:Wave on StringSubtopic:Equation/graph

⏱ 2mℹ️ Source: PYQ_2018

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