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A stretched string of length  1m fixed  at both ends, having a mass 5×10−4kg is under a tension of 20N. It is plucked at a point situated at 25 cm from one end. The stretched string could vibrate with a frequency of  
  • a)
    512 Hz
  • b)
    100 Hz
  • c)
    200 Hz
  • d)
    256 Hz
Correct answer is option 'C'. Can you explain this answer?
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A stretched string of length 1m fixed at both ends, having a mass 5...
At 25 cm, there will be antinode. So wire will vibrate in two loops  
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Most Upvoted Answer
A stretched string of length 1m fixed at both ends, having a mass 5...
Given information:

- Length of the string (L) = 1 m
- Mass of the string (m) = 5104 kg
- Tension in the string (T) = 20 N
- Distance of the plucked point from one end (x) = 25 cm = 0.25 m

Formula:

The frequency (f) of a vibrating string is given by the formula:

f = (1/2L) * sqrt(T/m)

Calculations:

Given:
- Length of the string (L) = 1 m
- Mass of the string (m) = 5104 kg
- Tension in the string (T) = 20 N

Using the formula:
f = (1/2L) * sqrt(T/m)

Substituting the given values:
f = (1/2 * 1) * sqrt(20/5104)

Simplifying the expression:
f = 0.5 * sqrt(20/5104)

Taking the square root:
f = 0.5 * 0.063

Evaluating the expression:
f = 0.0315

Converting the frequency to Hz:
f = 0.0315 Hz

Rounding off to the nearest whole number:
f ≈ 0 Hz

Therefore, the frequency of vibration of the stretched string is approximately 0 Hz.

However, it is important to note that the given answer options are all positive frequencies. It is possible that there is an error in the given question or answer options.
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A stretched string of length 1m fixed at both ends, having a mass 5×10−4kg is under a tension of 20N. It is plucked at a point situated at 25 cm from one end. The stretched string could vibrate with a frequency of a)512 Hzb)100 Hzc)200 Hzd)256 HzCorrect answer is option 'C'. Can you explain this answer?
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