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A String under a tension of 129.6 N produces 10 beats per sec when is vibrated along with a tuning frok. when tension in the string is increased to 160N, it sounds in unison with the same tuning frok. calculate the fundamental frequency of a tuning frok? (in Hz)
    Correct answer is '100'. Can you explain this answer?
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    A String under a tension of 129.6 N produces 10 beats per sec when is ...
    As here the tension in the wire is changed , so its fundamental frequency

    Now with increase in tension,
    (K√T) will increase & beats & decreasing to zero when 
    T = 160N (as union means frequencies are equal)
     ...(i)
     ...(ii)
    Substituting the value of ‘k’ from second equation in first

    f - 0.9 f = 10
    f = 100Hz
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    Most Upvoted Answer
    A String under a tension of 129.6 N produces 10 beats per sec when is ...
    As here the tension in the wire is changed , so its fundamental frequency

    Now with increase in tension,
    (K√T) will increase & beats & decreasing to zero when 
    T = 160N (as union means frequencies are equal)
     ...(i)
     ...(ii)
    Substituting the value of ‘k’ from second equation in first

    f - 0.9 f = 10
    f = 100Hz
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    Community Answer
    A String under a tension of 129.6 N produces 10 beats per sec when is ...
    Fundamental Frequency of a Tuning Fork

    Given:
    Tension in the string (T1) = 129.6 N
    Tension in the string (T2) = 160 N
    Number of beats per second (n) = 10

    To find:
    Fundamental frequency of the tuning fork (f)

    Formula:
    The frequency of the beats produced when two waves of slightly different frequencies interfere is given by the equation:

    f_beat = |f1 - f2|

    where f1 and f2 are the frequencies of the two waves.

    The fundamental frequency of the tuning fork can be found by using the equation:

    f = |f_string - f_tuning fork|

    where f_string is the frequency of the string and f_tuning fork is the frequency of the tuning fork.

    Understanding the Problem:

    In the given problem, the string is producing 10 beats per second when vibrated along with the tuning fork under a tension of 129.6 N. When the tension is increased to 160 N, the string and the tuning fork produce a unison sound.

    Solution:

    Step 1: Finding the frequency difference between the string and the tuning fork under a tension of 129.6 N.

    f_beat = n = 10 beats per second

    Since the frequency difference between the string and the tuning fork is equal to the number of beats per second, we can write:

    f_string - f_tuning fork = n
    f_string - f_tuning fork = 10

    Step 2: Finding the frequency difference between the string and the tuning fork under a tension of 160 N.

    f_beat = n = 0 beats per second

    Since there are no beats when the string and the tuning fork produce a unison sound, we can write:

    f_string - f_tuning fork = 0

    Step 3: Solving the equations to find the frequencies.

    Using the equations from steps 1 and 2, we can write:

    f_string - f_tuning fork = 10
    f_string - f_tuning fork = 0

    Subtracting the second equation from the first equation, we get:

    10 - 0 = 10

    So, the frequency difference between the string and the tuning fork is 10 Hz.

    Step 4: Finding the fundamental frequency of the tuning fork.

    Since the unison sound is produced when the frequency difference is 0, we can write:

    f_string - f_tuning fork = 0

    Substituting the value of f_string - f_tuning fork from step 3, we get:

    10 = 0

    Therefore, the fundamental frequency of the tuning fork is 10 Hz.

    However, in the question, the correct answer is given as 100 Hz. It seems there might be a mistake or discrepancy in the given information or answer. Please double-check the question or provide additional information if available.
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    A String under a tension of 129.6 N produces 10 beats per sec when is vibrated along with a tuning frok. when tension in the string is increased to 160N, it sounds in unison with the same tuning frok. calculate the fundamental frequency of a tuning frok? (in Hz)Correct answer is '100'. Can you explain this answer?
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    A String under a tension of 129.6 N produces 10 beats per sec when is vibrated along with a tuning frok. when tension in the string is increased to 160N, it sounds in unison with the same tuning frok. calculate the fundamental frequency of a tuning frok? (in Hz)Correct answer is '100'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A String under a tension of 129.6 N produces 10 beats per sec when is vibrated along with a tuning frok. when tension in the string is increased to 160N, it sounds in unison with the same tuning frok. calculate the fundamental frequency of a tuning frok? (in Hz)Correct answer is '100'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A String under a tension of 129.6 N produces 10 beats per sec when is vibrated along with a tuning frok. when tension in the string is increased to 160N, it sounds in unison with the same tuning frok. calculate the fundamental frequency of a tuning frok? (in Hz)Correct answer is '100'. Can you explain this answer?.
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