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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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Attempt All sub parts from each question.Tuning a piano wire: When we hear two frequencies which are very close to each other but not exactly equal then distinct waxing and waning of the intensity of the sound is heard. These are known as beats. The beat frequency is equal to the absolute value of the difference in frequency of the two waves. To tune a piano, a musician takes help of beat frequency. He plucks the string and tap a tuning fork at the same time. If the two sound sources— the piano string and the tuning fork, do not produce identical frequency then the musician hears detectable beats. Musician will then adjust the tension of the piano string and repeat the process until the beats are no longer be heard. Tension of the wire changes the frequency of the wire.As the piano string becomes more in tune with the tuning fork, the beat frequency reduces and approaches 0 Hz. When beats are no longer heard, the piano string is tuned to the tuning fork; i.e. they play the same frequency. The process allows the musician to match the strings' frequencies to the frequencies of a standardized set of tuning forks.Q. A musician while tuning a piano with a standard tuning fork of frequency 256 Hz, hears beats of 5 Hz. What he should do to achieve the tuning?

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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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