Question:
A stretched string of length 1 m fixed at both ends, when vibrated in one loop has a frequency of 200 Hz. It is now plucked at a point situated at 25 cm from one end. The stretched string would vibrate with a frequency of:
A stretched string of length 1 m fixed at both ends, when vibrated in one loop has a frequency of 200 Hz. It is now plucked at a point situated at 25 cm from one end. The stretched string would vibrate with a frequency of:
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The frequency of a string is inversely proportional to its length when other factors remain constant.
Updated On: Apr 1, 2025
- 100 Hz
- 200 Hz
- 400 Hz
- 800 Hz
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The Correct Option is C
Solution and Explanation
The frequency of a vibrating string is determined by the length of the string, the tension, and the mass per unit length.
The fundamental frequency is given by the equation: \[ f = \frac{v}{2L} \] where \( L \) is the length of the string, and \( v \) is the wave speed. When the string is plucked at a point 25 cm from one end, the effective length of the vibrating portion of the string is reduced, which increases the frequency.
Since the frequency is inversely proportional to the length of the vibrating portion, the frequency is doubled.
Thus, the correct answer is (c).
The fundamental frequency is given by the equation: \[ f = \frac{v}{2L} \] where \( L \) is the length of the string, and \( v \) is the wave speed. When the string is plucked at a point 25 cm from one end, the effective length of the vibrating portion of the string is reduced, which increases the frequency.
Since the frequency is inversely proportional to the length of the vibrating portion, the frequency is doubled.
Thus, the correct answer is (c).
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