JEE MAIN - Physics (2023 - 11th April Evening Shift - No. 22)
Explanation
To determine the fundamental frequency of the vibrating wire, we need to first find the tension (T) in the wire and the wave velocity (V) in the wire.
- Tension in the wire (T): We used Young's modulus (Y) to relate the stress and strain in the wire. The formula for stress is:
$$\text{stress} = Y \times \text{strain}$$
Here, the strain is the extension ($$\Delta L$$) divided by the original length (L):
$$\text{strain} = \frac{\Delta L}{L}$$
Now, the tension (T) in the wire is the product of stress and cross-sectional area (A):
$$T = \text{stress} \times A$$
Combining the above equations, we get the expression for tension:
$$T = \frac{Y \Delta L}{L} \times A$$
- Wave velocity in the wire (V): The linear mass density ($$\mu$$) of the wire is given by:
$$\mu = \frac{m}{L}$$
We need to find the ratio $$\frac{T}{\mu}$$, which represents the square of the wave velocity. Using the expressions for tension and linear mass density, we get:
$$\frac{T}{\mu} = \frac{Y \Delta L}{L} \times \frac{A}{m} = \frac{Y \Delta L}{L} \times \frac{1}{\rho}$$
Here, $$\rho$$ is the density of the wire material. Plugging in the given values, we find the value of $$\frac{T}{\mu}$$, which is:
$$\frac{T}{\mu} = 6.4 \times 10^3$$
Now, we find the wave velocity (V) by taking the square root of $$\frac{T}{\mu}$$:
$$V = \sqrt{T/\mu} = 80 \mathrm{~m/s}$$
- Fundamental frequency (f): Finally, we find the fundamental frequency of the vibrating wire using the formula:
$$f = \frac{V}{2L}$$
Plugging in the values, we get the fundamental frequency (f) as:
$$f = 80 \mathrm{~Hz}$$
So, the fundamental frequency of vibration in the wire is 80 Hz.
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