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? (a) 7.5 kHz (b) 5 kHz (c) 2.5 kHz (d) 10 kHz

Answer: (b). Clamped at the middle, both ends free: fundamental has L = (λ)/(2), so λ = 2 × 0.6 = 1.2 m. v = √(Y/ρ) = √9.27 × 10¹⁰2.7 × 10³ ≈ 5.86 × 10³ m/s f = (v)/(λ) = (5860)/(1.2) ≈ 4.9 × 10³ Hz So the answer is (b) 5 kHz.

A granite rod of 60 cm length is clamped at its middle point and is… | JEE Main 2018

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Speed of Waves

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 $\times$ 103 kg/m3 and its Young’s modulus is 9.27 $\times$ 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations?

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Answer: (b)

Clamped at the middle, both ends free: fundamental has $L = \dfrac{\lambda}{2}$ , so $\lambda = 2 \times 0.6 = 1.2\text{ m}$ .
$v = \sqrt{\dfrac{Y}{\rho}} = \sqrt{\dfrac{9.27 \times 10^{10}}{2.7 \times 10^{3}}} \approx 5.86 \times 10^{3}\text{ m/s}$
$f = \dfrac{v}{\lambda} = \dfrac{5860}{1.2} \approx 4.9 \times 10^{3}\text{ Hz}$
So the answer is (b) $5\text{ kHz}$ .

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