Assertion: If we see the oscillations of a stretched wire at higher overtone mode, frequency of oscillation increases but wavelength decreases.
Reason: From `v=f lambda, lambda prop 1/f` as `v` = constant.

Options

A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.
B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
C. If Assertion is true, but the Reason is false.
D. If Assertion is false but the Reason is true.

Explanation

To solve the question, we need to analyze the assertion and the reason provided:

**Assertion:** If we see the oscillations of a stretched wire at a higher overtone mode, the frequency of oscillation increases but the wavelength decreases.

**Reason:** From the relation \( v = f \lambda \), \( \lambda \) is proportional to \( \frac{1}{f} \) as \( v \) is constant.

### Step-by-Step Solution:

1. **Understanding Wave Properties:**
- In a stretched wire, waves can oscillate in different modes, known as harmonics or overtones. The fundamental frequency is the first harmonic, and the higher frequencies are called overtones.

2. **Analyzing the First Overtone:**
- For the first overtone (or second harmonic), the wave has one node at each end and one antinode in the middle. The wavelength (\( \lambda_1 \)) for the first overtone can be expressed as:
\[ \lambda_1 = \frac{2L}{1} = 2L \]
- Here, \( L \) is the length of the wire.

3. **Analyzing the Second Overtone:**
- For the second overtone (or third harmonic), there are two nodes and two antinodes. The wavelength (\( \lambda_2 \)) for the second overtone can be expressed as:
\[ \lambda_2 = \frac{2L}{3} \]
- As we can see, the wavelength decreases as we move to a higher overtone.

4. **Frequency Calculation:**
- The frequency (\( f \)) of a wave is inversely related to its wavelength when the wave speed (\( v \)) is constant:
\[ v = f \lambda \]
- Rearranging gives:
\[ f = \frac{v}{\lambda} \]
- Since \( v \) is constant for a given medium, as the wavelength decreases, the frequency must increase.

5. **Conclusion:**
- Therefore, as we move to higher overtone modes, the frequency of oscillation indeed increases while the wavelength decreases. This confirms the assertion.

6. **Evaluating the Reason:**
- The reason provided states that \( \lambda \) is proportional to \( \frac{1}{f} \) when \( v \) is constant. This is correct because it directly relates the frequency and wavelength in a wave equation.

7. **Final Statement:**
- Both the assertion and reason are true, and the reason correctly explains the assertion.

### Final Answer:
Both assertion and reason are true, and the reason is the correct explanation of the assertion.

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