A uniform wire of length I and mass M is stretched between two fixed points, keeping a tension F. A sound of frequency `mu` is impressed on it. Then the maximum vibrational energy is existing in the wire when `mu`

Options

A. `(1)/(2)sqrt((ML)/(F))`
B. `sqrt((FL)/(M))`
C. `2xxsqrt((FM)/(L))`
D. `(1)/(2)sqrt((F)/(ML))`

Answer

To solve the problem, we need to derive the expression for the frequency of the vibrating wire under tension. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We have a uniform wire of length \( L \) and mass \( M \) stretched between two fixed points with a tension \( F \). A sound of frequency \( \mu \) is impressed on it. We need to find the maximum vibrational energy in the wire when the frequency is \( \mu \).

### Step 2: Define Linear Density
First, we define the linear density \( \mu \) of the wire:
\[ \mu = \frac{M}{L} \]
where \( M \) is the mass of the wire and \( L \) is its length.

### Step 3: Use the Wave Equation for a Stretched Wire
The frequency of the fundamental mode of vibration of a stretched wire is given by the formula:
\[ \mu = \frac{1}{2L} \sqrt{\frac{F}{\mu}} \]
where \( F \) is the tension in the wire.

### Step 4: Rearranging the Formula
We can rearrange this formula to express \( \mu \):
\[ \mu = \frac{1}{2L} \sqrt{\frac{F}{\frac{M}{L}}} \]
This simplifies to:
\[ \mu = \frac{1}{2L} \sqrt{\frac{FL}{M}} \]

### Step 5: Simplifying Further
Now we can simplify the expression:
\[ \mu = \frac{1}{2} \sqrt{\frac{F}{M/L}} = \frac{1}{2} \sqrt{\frac{F}{\mu}} \]

### Step 6: Final Expression
Thus, we can write the expression for the frequency \( \mu \) as:
\[ \mu = \frac{1}{2} \sqrt{\frac{F}{M/L}} \]

### Step 7: Conclusion
The maximum vibrational energy in the wire occurs at this frequency \( \mu \).

### Final Answer
The maximum vibrational energy in the wire when \( \mu \) is:
\[ \mu = \frac{1}{2} \sqrt{\frac{F}{\frac{M}{L}}} \]