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 modules is 9.27 × 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ?

Question

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 modules is 9.27 × 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ? (A) 5 kHz (B) 2.5 kHz (C) 10 kHz (D) 7.5 kHz

Tardigrade Admin · Accepted Answer

f0=(V/2 L)=(1/2 L) √(Y/ρ) =(1/2 × 0.6) √(9.27 × 1010/2.7 × 103) =4.88 kHz ≈ 5 kHz

Q. 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 $2.7 \times 1 0^{3} k g / m^{3}$ and its Young's modules is $9.27 \times 1 0^{10} P a$ . What will be the fundamental frequency of the longitudinal vibrations ?

5 kHz

2.5 kHz

10 kHz

7.5 kHz

$f_{0} = \frac{V}{2 L} = \frac{1}{2 L} \frac{Y}{ρ}$
$= \frac{1}{2 \times 0.6} \frac{9.27 \times 1 0 ^{10}}{2.7 \times 1 0 ^{3}}$
$= 4.88 k Hz \approx 5 k Hz$

Q. A granite rod of 60cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7×103kg/m3 and its Young's modules is 9.27×1010Pa. What will be the fundamental frequency of the longitudinal vibrations ?