A long string is constructed by joining the ends of two shorter strings. The tension in the strings is the same but string I has 4 times the linear mass density of string II. When a sinusoidal wave passes from string I to string II:

Options

A. the frequency decreases by a factor of 4.
B. the frequency decreases by a factor of 2.
C. the wavelength decreases by a factor of 4.
D. the wavelength decreases by a factor of 2.

Answer

To solve the problem, we need to analyze the behavior of the waves as they pass from string I to string II, taking into account the linear mass densities and the tension in the strings.

### Step-by-Step Solution:

1. **Identify Given Information:**
- Let the linear mass density of string II be \( \mu \).
- Then, the linear mass density of string I is \( 4\mu \) (since it is 4 times that of string II).
- The tension \( T \) in both strings is the same.

2. **Determine Wave Speed in Each String:**
- The wave speed \( v \) in a string is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
- For string I (with linear mass density \( 4\mu \)):
\[ v_1 = \sqrt{\frac{T}{4\mu}} = \frac{1}{2} \sqrt{\frac{T}{\mu}} \]
- For string II (with linear mass density \( \mu \)):
\[ v_2 = \sqrt{\frac{T}{\mu}} \]

3. **Compare the Wave Speeds:**
- From the calculations:
\[ v_1 = \frac{1}{2} v_2 \]
- This means the wave speed in string I is half that of string II.

4. **Determine the Relationship Between Frequency and Wavelength:**
- The frequency \( f \) of a wave is related to its speed \( v \) and wavelength \( \lambda \) by the equation:
\[ v = f \lambda \]
- Rearranging gives:
\[ f = \frac{v}{\lambda} \]

5. **Calculate Frequencies in Both Strings:**
- Since the frequency remains constant when a wave travels from one medium to another, we can denote the frequency in both strings as \( f \).
- For string I:
\[ f_1 = \frac{v_1}{\lambda_1} \]
- For string II:
\[ f_2 = \frac{v_2}{\lambda_2} \]
- Since \( f_1 = f_2 = f \), we can write:
\[ \frac{v_1}{\lambda_1} = \frac{v_2}{\lambda_2} \]

6. **Express Wavelengths in Terms of Each Other:**
- Substituting \( v_1 \) and \( v_2 \):
\[ \frac{\frac{1}{2}v_2}{\lambda_1} = \frac{v_2}{\lambda_2} \]
- This simplifies to:
\[ \lambda_2 = 2\lambda_1 \]
- Thus, the wavelength in string II is twice that in string I.

### Conclusion:
- The frequency remains the same when the wave passes from string I to string II.
- The wavelength increases by a factor of 2.

### Final Answer:
- The frequency remains constant, and the wavelength increases by a factor of 2.