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?

We are given a granite rod whose total length is $$L = 60 \text{ cm} = 0.60 \text{ m}$$. The rod is clamped exactly at its mid-point, so the centre is rigidly fixed while both ends are free to move.

Because of the clamp, each half of the rod behaves like an independent bar of length $$\dfrac{L}{2} = \dfrac{0.60}{2} = 0.30 \text{ m}$$ which is fixed at one end (the clamped mid-point) and free at the other end.

For longitudinal vibrations of a bar with one end fixed and the other end free, the fundamental (lowest) mode has a node at the fixed end and an antinode at the free end. In such a mode the length of the bar equals one-quarter of the wavelength. We therefore write the relation

$$\text{Length of half-rod} = \dfrac{\lambda_1}{4}$$

Substituting $$0.30 \text{ m}$$ for the length, we obtain

$$0.30 = \dfrac{\lambda_1}{4} \quad\Longrightarrow\quad \lambda_1 = 4 \times 0.30 = 1.20 \text{ m}$$

The speed of longitudinal elastic waves in a solid is given by the formula

$$v = \sqrt{\dfrac{Y}{\rho}}$$

where $$Y = 9.27 \times 10^{10} \text{ Pa}$$ is Young’s modulus of granite, and $$\rho = 2.7 \times 10^{3} \text{ kg m}^{-3}$$ is its density.

We first evaluate the ratio inside the square root:

$$\dfrac{Y}{\rho} = \dfrac{9.27 \times 10^{10}}{2.7 \times 10^{3}} = \dfrac{9.27}{2.7} \times 10^{7} = 3.433\ \times 10^{7}$$

Taking the square root gives

$$v = \sqrt{3.433 \times 10^{7}} = \sqrt{3.433}\,\sqrt{10^{7}} = 1.852 \times 3162 \approx 5.86 \times 10^{3} \text{ m s}^{-1}$$

So the velocity of longitudinal waves in the granite rod is

$$v \approx 5.86 \times 10^{3} \text{ m s}^{-1}$$

The fundamental frequency $$f_1$$ is obtained from the basic wave relation $$v = f \lambda$$, hence

$$f_1 = \dfrac{v}{\lambda_1} = \dfrac{5.86 \times 10^{3}}{1.20} = 4.88 \times 10^{3} \text{ Hz} \approx 5.0 \text{ kHz}$$

Hence, the correct answer is Option B.