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 - 103 kg - m 3 and its Young-s modulus is 9-27 - 1010 Pa- What will be the fundamental frequency of the longitudinal vibrations-A- 10 kHzB- 7-5 kHzC- 5 kHzD- 2-5 kHz

F_0=1/2 ℓ√(Y/ρ)=1/2(0.6)√(9.27 × 10^10/2.7 × 10^3) =4.9 × 10^3Hz≃ 5 kHz ...

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Standard XII

Physics

String Fixed at Both Ends

A granite rod...

Question

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

10 k H z

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7.5 k H z

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5 k H z

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2.5 k H z

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Solution

The correct option is C $5 k H z$
$f_{0} = \frac{1}{2 ℓ} \sqrt{\frac{Y}{ρ}} = \frac{1}{2 (0.6)} \sqrt{\frac{9.27 \times 10^{10}}{2.7 \times 10^{3}}}$
$= 4.9 \times 10^{3} H z ≃ 5 k H z$

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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 2.7×103 kg/m3 and its Young's modulus is 9.27×1010 Pa. What will be the fundamental frequency of the longitudinal vibrations?

The correct option is C 5kHzf0=12ℓ√Yρ=12(0.6)√9.27×10102.7×103 =4.9×103Hz≃5kHz

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

The correct option is C $5 k H z$
$f_{0} = \frac{1}{2 ℓ} \sqrt{\frac{Y}{ρ}} = \frac{1}{2 (0.6)} \sqrt{\frac{9.27 \times 10^{10}}{2.7 \times 10^{3}}}$
$= 4.9 \times 10^{3} H z ≃ 5 k H z$