A light string is tied at one end to a fixed support and to a heavy string of equal length L at the other end A as shown in the figure ( Total length of both strings combined is 2L). A block of mass M is tied to the free end of heavy string. Mass per unit lenght of the string are `mu` and `16mu` and tensions is T . Find lowest positive value of frequency such that junction point A is a node.

Options

A. `(1)/(L)sqrt((T)/(mu))`
B. `(5)/(2L)sqrt((T)/(mu))`
C. `(3)/(2L)/sqrt((T)/(um))`
D. `(1)/(2L)sqrt((T)/(mu))`

Answer

To solve the problem, we need to analyze the setup of the strings and the block attached to the heavy string. The goal is to find the lowest positive frequency such that the junction point A is a node.

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a light string (mass per unit length = μ) and a heavy string (mass per unit length = 16μ).
- Both strings have the same length L.
- A block of mass M is attached to the free end of the heavy string.

2. **Identify the Conditions for a Node**:
- For point A to be a node, the wavelengths of the waves in both strings must satisfy certain conditions.
- The lengths of the strings can be expressed in terms of the wavelength (λ) as follows:
- For the light string: \( \frac{n_1 \lambda}{2} = L \) (where \( n_1 \) is the mode number for the light string)
- For the heavy string: \( \frac{n_2 \lambda}{2} = L \) (where \( n_2 \) is the mode number for the heavy string)

3. **Express Wavelength in Terms of Mode Numbers**:
- From the equations above, we can express the wavelength for both strings:
- \( \lambda = \frac{2L}{n_1} \) for the light string
- \( \lambda = \frac{2L}{n_2} \) for the heavy string

4. **Relate the Frequencies**:
- The frequency \( f \) is related to the wave speed \( v \) and wavelength \( \lambda \) by the equation:
\[ f = \frac{v}{\lambda} \]
- The wave speed in each string is given by:
- For the light string: \( v_1 = \sqrt{\frac{T}{\mu}} \)
- For the heavy string: \( v_2 = \sqrt{\frac{T}{16\mu}} \)

5. **Set Up the Frequency Equations**:
- For the light string:
\[ f_1 = \frac{v_1}{\lambda} = \frac{\sqrt{\frac{T}{\mu}}}{\frac{2L}{n_1}} = \frac{n_1}{2L} \sqrt{\frac{T}{\mu}} \]
- For the heavy string:
\[ f_2 = \frac{v_2}{\lambda} = \frac{\sqrt{\frac{T}{16\mu}}}{\frac{2L}{n_2}} = \frac{n_2}{2L} \sqrt{\frac{T}{16\mu}} \]

6. **Equate the Frequencies**:
- Since the frequencies must be equal at the node:
\[ \frac{n_1}{2L} \sqrt{\frac{T}{\mu}} = \frac{n_2}{2L} \sqrt{\frac{T}{16\mu}} \]
- Simplifying gives:
\[ n_1 \sqrt{16} = n_2 \implies 4n_1 = n_2 \]

7. **Find the Lowest Values of n1 and n2**:
- The lowest positive integer value for \( n_1 \) is 1, which gives \( n_2 = 4 \).

8. **Calculate the Frequency**:
- Using \( n_1 = 1 \) and \( n_2 = 4 \):
\[ f = \frac{n_1}{2L} \sqrt{\frac{T}{\mu}} = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

### Final Answer:
The lowest positive value of frequency \( f \) such that junction point A is a node is:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]