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A stretched string resonates with tuning fork frequency 512 Hz when length of the string is 0.5 m. The length of the string required to vibrate resonantly with a tuning fork of frequency 256 Hz would be [1993]
  • a)
    0.25 m
  • b)
    0.5 m
  • c)
    2 m
  • d)
    1 m
Correct answer is option 'D'. Can you explain this answer?
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A stretched string resonates with tuning fork frequency 512 Hz when le...

When f is halved, the length is doubled.
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A stretched string resonates with tuning fork frequency 512 Hz when le...
To find the length of the string required to vibrate resonantly with a tuning fork of frequency 256 Hz, we can use the formula for the fundamental frequency of a stretched string:

f = (1/2L) * sqrt(T/μ)

where:
f = frequency of vibration
L = length of the string
T = tension in the string
μ = linear mass density of the string

In this case, we know that when the length of the string is 0.5 m, it resonates with a tuning fork frequency of 512 Hz. Let's denote this as Case 1.

Case 1:
Length of string (L1) = 0.5 m
Tuning fork frequency (f1) = 512 Hz

We can rearrange the formula to solve for T:

T = (f^2) * μ * (2L)^2

Now, we need to find the tension in the string for Case 1.

Next, we can use the same formula to find the length of the string (L2) required to resonate with a tuning fork frequency of 256 Hz. Let's denote this as Case 2.

Case 2:
Tuning fork frequency (f2) = 256 Hz

We can rearrange the formula to solve for L:

L = (1/2) * sqrt(T/μ) * (1/f)

Now, we need to substitute the tension value from Case 1 into this formula to find the length of the string for Case 2.

By substituting the values into the formulas and performing the calculations, we find that the length of the string required to vibrate resonantly with a tuning fork of frequency 256 Hz is 1 m, which corresponds to option D.
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A stretched string resonates with tuning fork frequency 512 Hz when length of the string is 0.5 m. The length of the string required to vibrate resonantly with a tuning fork of frequency 256 Hz would be [1993]a)0.25 mb)0.5 mc)2 md)1 mCorrect answer is option 'D'. Can you explain this answer?
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A stretched string resonates with tuning fork frequency 512 Hz when length of the string is 0.5 m. The length of the string required to vibrate resonantly with a tuning fork of frequency 256 Hz would be [1993]a)0.25 mb)0.5 mc)2 md)1 mCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A stretched string resonates with tuning fork frequency 512 Hz when length of the string is 0.5 m. The length of the string required to vibrate resonantly with a tuning fork of frequency 256 Hz would be [1993]a)0.25 mb)0.5 mc)2 md)1 mCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stretched string resonates with tuning fork frequency 512 Hz when length of the string is 0.5 m. The length of the string required to vibrate resonantly with a tuning fork of frequency 256 Hz would be [1993]a)0.25 mb)0.5 mc)2 md)1 mCorrect answer is option 'D'. Can you explain this answer?.
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