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, we need to determine the wavelength of the wave at the top end of the string given the wavelength at the bottom end and the relationship between tension and wave speed in a string.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Mass of the suspended object (m1) = 2 kg
- Mass of the string (m2) = 6 kg
- Wavelength at the bottom of the string (λ1) = 6 cm

2. **Calculate Tension at the Bottom of the String (T1):**
- The tension at the bottom of the string (T1) is equal to the weight of the suspended mass.
\[ T1 = m1 \cdot g = 2 \, \text{kg} \cdot g \]
(where g is the acceleration due to gravity, approximately \(9.81 \, \text{m/s}^2\), but we can keep it as g for our calculations).

3. **Calculate Tension at the Top of the String (T2):**
- The tension at the top of the string (T2) is equal to the weight of the suspended mass plus the weight of the string.
\[ T2 = (m1 + m2) \cdot g = (2 \, \text{kg} + 6 \, \text{kg}) \cdot g = 8 \, \text{kg} \cdot g \]

4. **Relate Wave Speeds and Wavelengths:**
- The wave speed in a string is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where T is the tension and μ is the mass per unit length of the string. Since we are not given the mass per unit length, we can use the relationship between the wave speeds and wavelengths:
\[ \frac{v1}{v2} = \frac{\lambda1}{\lambda2} \]
where \(v1\) and \(v2\) are the wave speeds at the bottom and top of the string, respectively.

5. **Express Wave Speeds in Terms of Tension:**
- Since the wave speed is proportional to the square root of tension, we can write:
\[ \frac{v1}{v2} = \sqrt{\frac{T1}{T2}} \]

6. **Substituting Values:**
- From the previous steps:
\[ \frac{v1}{v2} = \sqrt{\frac{2g}{8g}} = \sqrt{\frac{2}{8}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

7. **Relate Wavelengths:**
- Now substituting this into the wavelength relationship:
\[ \frac{\lambda1}{\lambda2} = \frac{1}{2} \]
Rearranging gives:
\[ \lambda2 = \lambda1 \cdot 2 \]

8. **Calculate Wavelength at the Top of the String:**
- Substitute λ1 = 6 cm:
\[ \lambda2 = 6 \, \text{cm} \cdot 2 = 12 \, \text{cm} \]

### Final Answer:
The wavelength of the wave at the top end of the string is **12 cm**.