Question:
If the length and diameter of a wire are decreased, then for the same tension the natural frequency of stretched wire will
If the length and diameter of a wire are decreased, then for the same tension the natural frequency of stretched wire will
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When analyzing the frequency of a stretched wire, remember that frequency increases when the length decreases and the diameter decreases because the linear density decreases.
Updated On: Feb 18, 2026
- decrease.
- not change.
- become zero.
- increase.
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the relationship.
The natural frequency \( f \) of a stretched wire is given by: \[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \] where \( L \) is the length of the wire, \( T \) is the tension, and \( \mu \) is the linear density of the wire. The linear density \( \mu \) is related to the mass per unit length and cross-sectional area \( A \) of the wire: \[ \mu = \frac{m}{L} = \frac{\rho A}{L} \] Step 2: Analyzing the effects of decreasing length and diameter.
Decreasing the length \( L \) and diameter \( D \) (which decreases the area \( A \)) increases the natural frequency. Since the tension is the same, the frequency increases because both \( L \) and \( A \) appear in the formula for frequency.
Step 3: Conclusion.
Thus, when the length and diameter are decreased, the natural frequency of the stretched wire increases, corresponding to option (D).
The natural frequency \( f \) of a stretched wire is given by: \[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \] where \( L \) is the length of the wire, \( T \) is the tension, and \( \mu \) is the linear density of the wire. The linear density \( \mu \) is related to the mass per unit length and cross-sectional area \( A \) of the wire: \[ \mu = \frac{m}{L} = \frac{\rho A}{L} \] Step 2: Analyzing the effects of decreasing length and diameter.
Decreasing the length \( L \) and diameter \( D \) (which decreases the area \( A \)) increases the natural frequency. Since the tension is the same, the frequency increases because both \( L \) and \( A \) appear in the formula for frequency.
Step 3: Conclusion.
Thus, when the length and diameter are decreased, the natural frequency of the stretched wire increases, corresponding to option (D).
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