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 xx10^(3) kg//m^(3)` and its Young’s modulus is `9.27 xx10^(10)` Pa. What will be the fundamental frequency of the longitudinal vibrations ?

Options

A. (a)5kHz
B. (b)2.5kHz
C. (c)10kHz
D. (d)7.5kHz

Explanation

To find the fundamental frequency of the longitudinal vibrations of the granite rod, we can follow these steps:

### Step 1: Convert Length to Meters
The length of the granite rod is given as 60 cm. We need to convert this to meters.
\[ L = 60 \, \text{cm} = \frac{60}{100} \, \text{m} = 0.6 \, \text{m} \]

### Step 2: Identify Given Values
We have the following values:
- Density of granite, \(\rho = 2.7 \times 10^3 \, \text{kg/m}^3\)
- Young’s modulus, \(Y = 9.27 \times 10^{10} \, \text{Pa}\)

### Step 3: Calculate the Velocity of Sound in Granite
The velocity of sound in a material can be calculated using the formula:
\[ v = \sqrt{\frac{Y}{\rho}} \]
Substituting the values:
\[ v = \sqrt{\frac{9.27 \times 10^{10}}{2.7 \times 10^{3}}} \]
Calculating this gives:
\[ v = \sqrt{3.4333 \times 10^{7}} \approx 5.85 \times 10^{3} \, \text{m/s} \]

### Step 4: Calculate the Fundamental Frequency
The fundamental frequency \(f\) for a rod clamped at its middle point is given by:
\[ f = \frac{v}{2L} \]
Substituting the values we have:
\[ f = \frac{5.85 \times 10^{3}}{2 \times 0.6} \]
Calculating this gives:
\[ f = \frac{5.85 \times 10^{3}}{1.2} \approx 4.875 \times 10^{3} \, \text{Hz} \approx 4.88 \, \text{kHz} \]

### Final Answer
The fundamental frequency of the longitudinal vibrations of the granite rod is approximately **4.88 kHz**.

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