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Step-by-Step Solution
Step 1: Identify the physical situation
We have a granite rod of total length 60 cm. It is clamped at its middle point, effectively creating two segments of rod, each of length L' = 0.30 m. Each segment behaves like a rod that is fixed at one end (the clamp) and free at the other end (the rod’s free end) for longitudinal vibrations.
Step 2: Recall the formula for the fundamental frequency of a rod fixed at one end and free at the other
For a rod of length L (with one end fixed and the other end free), the fundamental frequency of longitudinal vibrations is given by:
$ f = \frac{1}{4L} \sqrt{\frac{Y}{\rho}}, $
where:
$ Y $ = Young’s modulus of the material of the rod.
$ \rho $ = density of the material of the rod.
$ L $ = length of the rod segment (in this problem, L' = 0.30 m).
Step 3: Substitute the given values
We are given:
$ Y = 9.27 \times 10^{10} \,\text{Pa} $
$ \rho = 2.7 \times 10^{3} \,\text{kg/m}^3 $
$ L' = 0.30 \,\text{m} $ (each half of the rod)
First, calculate the wave speed for longitudinal waves in the rod:
$ v = \sqrt{\frac{Y}{\rho}} = \sqrt{\frac{9.27 \times 10^{10}}{2.7 \times 10^{3}}}.
$
This value is approximately in the range of a few thousand meters per second.
Step 4: Compute the fundamental frequency
Using
$ f = \frac{1}{4L'} \sqrt{\frac{Y}{\rho}}, $
we set $ L' = 0.30\,\text{m} $:
$ f = \frac{1}{4 \times 0.30} \sqrt{\frac{9.27 \times 10^{10}}{2.7 \times 10^{3}}}.
$
Numerically, this comes out to approximately 5 kHz.
Step 5: Conclude the answer
The fundamental frequency of the granite rod, clamped at the middle and set into longitudinal vibrations, is 5 kHz.
