A mass of 2kg suspended by string of mass 6kg. A wave of wavelength 6 cm is produced at bottom of string. Wavelength of wave at top end of string will be

Options

A. 6 cm
B. 18 cm
C. 12 cm
D. 24 cm

Answer

To solve the problem step by step, we need to analyze the situation involving the mass, the string, and the wave properties.

### Step 1: Understand the system
We have a mass \( m = 2 \, \text{kg} \) suspended by a string of mass \( M = 6 \, \text{kg} \). A wave of wavelength \( \lambda_A = 6 \, \text{cm} \) is produced at the bottom of the string.

### Step 2: Calculate the tension at the bottom of the string
At the bottom of the string (point A), the tension \( T_A \) is equal to the weight of the suspended mass:
\[ T_A = m \cdot g = 2 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 19.6 \, \text{N} \approx 20 \, \text{N} \]

### Step 3: Calculate the tension at the top of the string
At the top of the string (point B), the tension \( T_B \) must support both the weight of the mass and the weight of the string itself. The total weight supported at point B is:
\[ \text{Total weight} = m \cdot g + M \cdot g = (2 \, \text{kg} + 6 \, \text{kg}) \cdot g = 8 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 78.4 \, \text{N} \approx 80 \, \text{N} \]

### Step 4: Relate the velocities and tensions
The wave velocity in a string is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension and \( \mu \) is the mass per unit length. Since the mass of the string is distributed along its length, we can express the mass per unit length as \( \mu = \frac{M}{L} \), where \( L \) is the length of the string.

### Step 5: Relate the wavelengths using the velocities
Since the frequency \( f \) of the wave remains constant, we can relate the velocities and wavelengths at points A and B:
\[ \frac{v_B}{v_A} = \frac{\lambda_B}{\lambda_A} \]
Using the relationship between velocity and tension:
\[ \frac{v_B}{v_A} = \sqrt{\frac{T_B}{T_A}} \]

### Step 6: Substitute the tensions
Substituting the values of tensions:
\[ \frac{v_B}{v_A} = \sqrt{\frac{T_B}{T_A}} = \sqrt{\frac{80 \, \text{N}}{20 \, \text{N}}} = \sqrt{4} = 2 \]

### Step 7: Calculate the wavelength at the top of the string
Now, substituting into the wavelength relationship:
\[ \frac{\lambda_B}{\lambda_A} = 2 \implies \lambda_B = 2 \cdot \lambda_A \]
Given \( \lambda_A = 6 \, \text{cm} \):
\[ \lambda_B = 2 \cdot 6 \, \text{cm} = 12 \, \text{cm} \]

### Final Answer
The wavelength of the wave at the top of the string is \( \lambda_B = 12 \, \text{cm} \).

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