An object of specific gravity `rho` is hung from a thin copper wire. The fundamental frequency for transverse standaing waves in the wire is 400 Hz. The object is immersed in water so that one third of its volume is submerged. What is the new fundamental frequency?

Options

A. `200sqrt((rho-1)/(3rho))Hz`
B. `400sqrt((3rho-1)/(3rho))Hz`
C. `400sqrt((rho-1)/(3rho))Hz`
D. `200sqrt((3rho-1)/(3rho))Hz`

Answer

To solve the problem, we need to find the new fundamental frequency of a wire when an object is partially submerged in water. We will follow these steps:

### Step 1: Understand the initial conditions
The initial frequency of the wire when the object is hanging freely is given as:
\[ f = 400 \, \text{Hz} \]

The tension in the wire when the object is hanging in air is:
\[ T = mg \]
where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity.

### Step 2: Express mass in terms of volume and density
The mass \( m \) can be expressed as:
\[ m = V \cdot \rho \]
where \( V \) is the volume of the object and \( \rho \) is the specific gravity of the object (density of the object divided by the density of water).

Thus, the tension can be rewritten as:
\[ T = V \cdot \rho \cdot g \]

### Step 3: Write the formula for frequency
The fundamental frequency \( f \) of a wire is given by:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where \( L \) is the length of the wire and \( \mu \) is the mass per unit length of the wire.

### Step 4: Analyze the submerged case
When the object is submerged in water with one-third of its volume submerged, the buoyant force \( F_b \) acting on the object is:
\[ F_b = \frac{1}{3} V \cdot g \]
The new tension \( T' \) in the wire when the object is submerged is:
\[ T' = mg - F_b \]
Substituting the expressions for \( m \) and \( F_b \):
\[ T' = V \cdot \rho \cdot g - \frac{1}{3} V \cdot g \]
\[ T' = Vg \left( \rho - \frac{1}{3} \right) \]

### Step 5: Write the new frequency equation
The new frequency \( f' \) can be expressed as:
\[ f' = \frac{1}{2L} \sqrt{\frac{T'}{\mu}} \]
Substituting \( T' \):
\[ f' = \frac{1}{2L} \sqrt{\frac{Vg \left( \rho - \frac{1}{3} \right)}{\mu}} \]

### Step 6: Relate the new frequency to the original frequency
We can relate \( f' \) to \( f \) by dividing the new frequency by the original frequency:
\[ \frac{f'}{f} = \sqrt{\frac{T'}{T}} \]
Substituting \( T \) and \( T' \):
\[ \frac{f'}{400} = \sqrt{\frac{Vg \left( \rho - \frac{1}{3} \right)}{Vg \rho}} \]
The \( Vg \) terms cancel out:
\[ \frac{f'}{400} = \sqrt{\frac{\rho - \frac{1}{3}}{\rho}} \]

### Step 7: Solve for the new frequency
Now, we can express \( f' \):
\[ f' = 400 \cdot \sqrt{\frac{\rho - \frac{1}{3}}{\rho}} \]

### Final Answer
Thus, the new fundamental frequency when the object is partially submerged is:
\[ f' = 400 \cdot \sqrt{\frac{3\rho - 1}{3\rho}} \]