QJEEM23Ph04e9: JEE Main 2023 Physics - Class 11 JEEM23Ph04e9 Solution: A wire of density is stretched between two clamps apart. The…

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.

1. 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:

$T = \frac{Y \Delta L}{L} \times A$

1. 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 $\frac{T}{\mu}$ :

$V = \sqrt{T/\mu} = 80 \mathrm{~m/s}$

1. 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.