To solve the problem regarding the frequency of a vibrating sonometer wire when the tension is doubled, we need to understand the relationship between frequency, tension, length, and mass per unit length of the wire. ### Step-by-Step Explanation 1. **Understanding the Basics**: The frequency of a vibrating wire (or string) is given by the formula: \[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \] where: - \( f \) is the frequency of vibration, - \( L \) is the length of the wire, - \( T \) is the tension in the wire, - \( \mu \) is the mass per unit length of the wire. 2. **Initial Conditions**: Let's denote the initial frequency as \( f_0 \). According to the formula, we can express this as: \[ f_0 = \frac{1}{2L} \sqrt{\frac{T_0}{\mu}} \] where \( T_0 \) is the initial tension. 3. **Doubling the Tension**: Now, if the tension is doubled, we have: \[ T = 2T_0 \] We need to find the new frequency \( f \) when the tension is \( 2T_0 \): \[ f = \frac{1}{2L} \sqrt{\frac{2T_0}{\mu}} \] 4. **Calculating the New Frequency**: We can simplify the expression for the new frequency: \[ f = \frac{1}{2L} \sqrt{2} \sqrt{\frac{T_0}{\mu}} = \sqrt{2} \cdot \frac{1}{2L} \sqrt{\frac{T_0}{\mu}} = \sqrt{2} \cdot f_0 \] Thus, the new frequency \( f \) is: \[ f = f_0 \sqrt{2} \] 5. **Final Answer**: Therefore, the new frequency of vibration when the tension is doubled is: \[ f = f_0 \sqrt{2} \] This corresponds to option **D**. ### Why Other Options Are Incorrect - **Option A: \( f_0/2 \)**: This option suggests that the frequency decreases to half. However, increasing the tension actually increases the frequency, not decreases it. - **Option B: \( 2f_0 \)**: This option implies that the frequency doubles. While increasing tension does increase frequency, it does not double it; it increases it by a factor of \( \sqrt{2} \). - **Option C: \( f_0/\sqrt{2} \)**: This option suggests that the frequency decreases. Again, this is incorrect because increasing tension leads to an increase in frequency. ### Common Pitfalls - **Misunderstanding the Relationship**: Students often confuse how tension affects frequency. Remember that frequency increases with tension, but the relationship is not linear; it is proportional to the square root of the tension. - **Forgetting Constants**: Ensure that when applying the formula, you keep the length and mass per unit length constant, as they are not changing in this scenario. ### Revision Summary - The frequency of a vibrating wire is given by \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \). - Doubling the tension increases the frequency by a factor of \( \sqrt{2} \). - The new frequency when tension is doubled is \( f = f_0 \sqrt{2} \). - The correct answer is option **D**: \( f_0 \sqrt{2} \).