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 ?

Waves PYQ — JEE Main 2018 · Shift 0 · Q45 with Solution

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Correct answer: B

Wave speed for longitudinal vibrations in a rod

For longitudinal waves in a solid rod,

v = \sqrt{\frac{Y}{\rho}}

where:

$Y = 9.27 \times 10^{10}\,\text{Pa}$
$\rho = 2.7 \times 10^3\,\text{kg/m}^3$

So,

v = \sqrt{\frac{9.27 \times 10^{10}}{2.7 \times 10^3}}

v = \sqrt{3.433\ldots \times 10^7}

v \approx 5.86 \times 10^3\,\text{m/s}

Condition for fundamental mode

The rod is clamped at the middle, so the middle point is a displacement node. For the fundamental longitudinal mode, each half of the rod vibrates like a rod fixed at one end and free at the other.

Thus,

\frac{\lambda}{4} = \frac{L}{2}

where total rod length is

L = 60\,\text{cm} = 0.60\,\text{m}

Hence,

\lambda = 2L = 1.2\,\text{m}

Frequency calculation

Using

f = \frac{v}{\lambda}

f = \frac{5.86 \times 10^3}{1.2}

f \approx 4.88 \times 10^3\,\text{Hz}

f \approx 4.9\,\text{kHz}

So the nearest option is:

\boxed{5\,\text{kHz}}

Option check

A: $7.5\,$ kHz $\;\times$
B: $5\,$ kHz $\;\checkmark$
C: $2.5\,$ kHz $\;\times$
D: $10\,$ kHz $\;\times$

Therefore, the correct answer is Option B.

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