A tuning fork is in unison with 1.0 m length of sonometer wire. When the stretching weights are immersed in water, the length of the wire in unison with the same tuning fork is 0.934 m. Calculate the density of the material of the weights ?

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between frequency and length of the wire
The frequency of a vibrating string (or wire) is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{W}{m}} \]
where:
- \( f \) is the frequency,
- \( L \) is the length of the wire,
- \( W \) is the weight (force due to gravity on the weights),
- \( m \) is the mass of the wire.

### Step 2: Set up the equations for the two lengths
Given:
- Length \( L_1 = 1.0 \, \text{m} \)
- Length \( L_2 = 0.934 \, \text{m} \)

Since the tuning fork is in unison with both lengths, we can set up two equations using the frequency formula:

1. For \( L_1 \):
\[ f_1 = \frac{1}{2L_1} \sqrt{\frac{W_1}{m}} \]

2. For \( L_2 \):
\[ f_2 = \frac{1}{2L_2} \sqrt{\frac{W_2}{m}} \]

### Step 3: Equate the frequencies
Since \( f_1 = f_2 \), we can set the two equations equal to each other:
\[ \frac{1}{2L_1} \sqrt{\frac{W_1}{m}} = \frac{1}{2L_2} \sqrt{\frac{W_2}{m}} \]

### Step 4: Simplify the equation
Cancel out the common terms:
\[ \frac{1}{L_1} \sqrt{W_1} = \frac{1}{L_2} \sqrt{W_2} \]

Rearranging gives:
\[ \frac{\sqrt{W_1}}{\sqrt{W_2}} = \frac{L_1}{L_2} \]

### Step 5: Substitute the known lengths
Substituting \( L_1 = 1.0 \, \text{m} \) and \( L_2 = 0.934 \, \text{m} \):
\[ \frac{\sqrt{W_1}}{\sqrt{W_2}} = \frac{1.0}{0.934} \]

### Step 6: Square both sides
Squaring both sides gives:
\[ \frac{W_1}{W_2} = \left(\frac{1.0}{0.934}\right)^2 \]
Calculating this:
\[ \frac{W_1}{W_2} \approx 1.146 \]

### Step 7: Relate the weights to density
The density \( \rho \) of the weights can be calculated using the formula:
\[ \rho = \frac{W}{V} \]
where \( V \) is the volume of the weights. The volume can be expressed in terms of the weight and density of water (since the weights are immersed in water):
\[ V = \frac{W}{\rho_{water}} \quad \text{(where } \rho_{water} \approx 1000 \, \text{kg/m}^3\text{)} \]

### Step 8: Calculate the density of the weights
Using the relationship between the weights:
\[ \text{Let } W_1 = W \text{ and } W_2 = \frac{W}{1.146} \]
The relative density of the weights can be expressed as:
\[ \text{Relative Density} = \frac{W_1}{W_1 - W_2} = \frac{W}{W - \frac{W}{1.146}} = \frac{1}{1 - \frac{1}{1.146}} \approx 7.849 \]

### Step 9: Convert to kg/m³
To convert from grams per cubic meter to kilograms per cubic meter:
\[ \rho \approx 7.849 \times 10^3 \, \text{kg/m}^3 \]

### Final Answer
The density of the material of the weights is approximately:
\[ \rho \approx 7849 \, \text{kg/m}^3 \]

---