Two uniform wire are vibrating simultaneously in their fundamental modes. The tensions, lengths, diameters, and the densities of the two wires are in the ratios, `8:1 , 36:35 , 4:1 , 1:2` respectively. If the note of higher pitch has a frequency of 360Hz the number of beats produced per second is

Options

A. 5
B. 10
C. 15
D. 20

Answer

To solve the problem, we need to find the frequencies of two vibrating wires and then calculate the number of beats produced per second. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the relationship between frequency and the parameters of the wires
The fundamental frequency \( \nu \) of a wire can be expressed as:
\[ \nu = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( L \) = length of the wire
- \( T \) = tension in the wire
- \( \mu \) = mass per unit length of the wire

The mass per unit length \( \mu \) can be calculated as:
\[ \mu = \frac{\pi d^2}{4} \rho \]
where:
- \( d \) = diameter of the wire
- \( \rho \) = density of the wire

### Step 2: Set up the ratios based on the problem statement
Given ratios:
- Tensions \( T_1 : T_2 = 8 : 1 \)
- Lengths \( L_1 : L_2 = 36 : 35 \)
- Densities \( \rho_1 : \rho_2 = 1 : 2 \)
- Diameters \( d_1 : d_2 = 4 : 1 \)

### Step 3: Write the frequency ratio
Using the ratios, we can express the ratio of the frequencies \( \nu_1 \) and \( \nu_2 \):
\[ \frac{\nu_1}{\nu_2} = \frac{L_2}{L_1} \cdot \sqrt{\frac{T_1}{T_2}} \cdot \frac{\rho_2}{\rho_1} \cdot \left(\frac{d_2^2}{d_1^2}\right) \]

### Step 4: Substitute the ratios into the frequency ratio
Substituting the given ratios into the equation:
\[ \frac{\nu_1}{\nu_2} = \frac{35}{36} \cdot \sqrt{\frac{8}{1}} \cdot \frac{2}{1} \cdot \left(\frac{1^2}{4^2}\right) \]

### Step 5: Simplify the expression
Calculating each part:
- \( \sqrt{\frac{8}{1}} = \sqrt{8} = 2\sqrt{2} \)
- \( \frac{1^2}{4^2} = \frac{1}{16} \)

Putting it all together:
\[ \frac{\nu_1}{\nu_2} = \frac{35}{36} \cdot 2\sqrt{2} \cdot 2 \cdot \frac{1}{16} \]
\[ = \frac{35 \cdot 4\sqrt{2}}{36 \cdot 16} = \frac{35\sqrt{2}}{144} \]

### Step 6: Calculate \( \nu_2 \) given \( \nu_1 \)
Given that the higher pitch frequency \( \nu_1 = 360 \, \text{Hz} \):
\[ \nu_2 = \nu_1 \cdot \frac{36}{35\sqrt{2}} = 360 \cdot \frac{36}{35\sqrt{2}} \]
Calculating \( \nu_2 \):
\[ \nu_2 \approx 350 \, \text{Hz} \]

### Step 7: Calculate the beat frequency
The beat frequency is given by the absolute difference between the two frequencies:
\[ \text{Beat frequency} = |\nu_1 - \nu_2| = |360 - 350| = 10 \, \text{Hz} \]

### Final Answer
The number of beats produced per second is **10 Hz**.

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