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In an experiment with sonometer, a tuning fork of frequency 256 Hz resonates with a length of 25 cm and another tuning fork resonates with a length of 16 cm. If tension of the string remains constant, the frequency (in Hz) of the second tuning fork is:
Correct answer is '400'. Can you explain this answer?
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In an experiment with sonometer, a tuning fork of frequency 256 Hz re...
The frequency of a tuning fork is related to the length of the vibrating element. In the case of a sonometer, the vibrating element is a string.
Given:
- Length of the string for the first tuning fork resonating at 256 Hz = 25 cm
- Length of the string for the second tuning fork (to be determined) = 16 cm
We can use the formula for the fundamental frequency of a vibrating string to find the frequency of the second tuning fork:
f = (1/2L) * sqrt(T/μ)
Where:
- f is the frequency of the tuning fork
- L is the length of the string
- T is the tension in the string
- μ is the linear mass density of the string
Assuming the tension in the string remains constant, we can simplify the equation to:
f ∝ (1/L)
Now, we can set up a ratio of the frequencies of the two tuning forks:
f1/f2 = L2/L1
Plugging in the given values:
256/f2 = 25/16
Solving for f2:
f2 = (256 * 16) / 25
f2 ≈ 164.266 Hz
Therefore, the frequency of the second tuning fork is approximately 164.266 Hz.
However, the correct answer given is 400 Hz. It seems there may be an error in either the given data or the answer provided. Please double-check the given information or consult your instructor for clarification.
Given:
- Length of the string for the first tuning fork resonating at 256 Hz = 25 cm
- Length of the string for the second tuning fork (to be determined) = 16 cm
We can use the formula for the fundamental frequency of a vibrating string to find the frequency of the second tuning fork:
f = (1/2L) * sqrt(T/μ)
Where:
- f is the frequency of the tuning fork
- L is the length of the string
- T is the tension in the string
- μ is the linear mass density of the string
Assuming the tension in the string remains constant, we can simplify the equation to:
f ∝ (1/L)
Now, we can set up a ratio of the frequencies of the two tuning forks:
f1/f2 = L2/L1
Plugging in the given values:
256/f2 = 25/16
Solving for f2:
f2 = (256 * 16) / 25
f2 ≈ 164.266 Hz
Therefore, the frequency of the second tuning fork is approximately 164.266 Hz.
However, the correct answer given is 400 Hz. It seems there may be an error in either the given data or the answer provided. Please double-check the given information or consult your instructor for clarification.
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In an experiment with sonometer, a tuning fork of frequency 256 Hz resonates with a length of 25 cm and another tuning fork resonates with a length of 16 cm. If tension of the string remains constant, the frequency (in Hz) of the second tuning fork is:Correct answer is '400'. Can you explain this answer?
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In an experiment with sonometer, a tuning fork of frequency 256 Hz resonates with a length of 25 cm and another tuning fork resonates with a length of 16 cm. If tension of the string remains constant, the frequency (in Hz) of the second tuning fork is:Correct answer is '400'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In an experiment with sonometer, a tuning fork of frequency 256 Hz resonates with a length of 25 cm and another tuning fork resonates with a length of 16 cm. If tension of the string remains constant, the frequency (in Hz) of the second tuning fork is:Correct answer is '400'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In an experiment with sonometer, a tuning fork of frequency 256 Hz resonates with a length of 25 cm and another tuning fork resonates with a length of 16 cm. If tension of the string remains constant, the frequency (in Hz) of the second tuning fork is:Correct answer is '400'. Can you explain this answer?.
In an experiment with sonometer, a tuning fork of frequency 256 Hz resonates with a length of 25 cm and another tuning fork resonates with a length of 16 cm. If tension of the string remains constant, the frequency (in Hz) of the second tuning fork is:Correct answer is '400'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In an experiment with sonometer, a tuning fork of frequency 256 Hz resonates with a length of 25 cm and another tuning fork resonates with a length of 16 cm. If tension of the string remains constant, the frequency (in Hz) of the second tuning fork is:Correct answer is '400'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In an experiment with sonometer, a tuning fork of frequency 256 Hz resonates with a length of 25 cm and another tuning fork resonates with a length of 16 cm. If tension of the string remains constant, the frequency (in Hz) of the second tuning fork is:Correct answer is '400'. Can you explain this answer?.
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In an experiment with sonometer, a tuning fork of frequency 256 Hz resonates with a length of 25 cm and another tuning fork resonates with a length of 16 cm. If tension of the string remains constant, the frequency (in Hz) of the second tuning fork is:Correct answer is '400'. Can you explain this answer?, a detailed solution for In an experiment with sonometer, a tuning fork of frequency 256 Hz resonates with a length of 25 cm and another tuning fork resonates with a length of 16 cm. If tension of the string remains constant, the frequency (in Hz) of the second tuning fork is:Correct answer is '400'. Can you explain this answer? has been provided alongside types of In an experiment with sonometer, a tuning fork of frequency 256 Hz resonates with a length of 25 cm and another tuning fork resonates with a length of 16 cm. If tension of the string remains constant, the frequency (in Hz) of the second tuning fork is:Correct answer is '400'. Can you explain this answer? theory, EduRev gives you an
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