Replacing an object attached to a spring with an object having 14 the original mass will change the frequency of oscillation of the system b

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Replacing an object attached to a spring with an object having 14 the original mass will change the frequency of oscillation of the system by a factor of

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Ngọc Hoa 5 years 2021-08-05T02:15:43+00:00 1 Answers 42 views 0

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  1. kimchi2
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    2021-08-05T02:16:44+00:00
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    Answer:

    The frequency changes by a factor of  0.27.

    Explanation:

    The frequency of an object with mass m attached to a spring is given as

    f = \frac{1}{2\pi } \sqrt{\frac{k}{m} }

    where f is the frequency

    k is the spring constant of the spring

    m is the mass of the substance on the spring.

    If the mass of the system is increased by 14 means the new frequency becomes

    f_{n} = \frac{1}{2\pi } \sqrt{\frac{k}{14m} }

    simplifying, we have

    f_{n} = \frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }

    f_{n} = \frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }

    if we divide this final frequency by the original frequency, we’ll have

    ==> \frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }  ÷  \frac{1}{2\pi } \sqrt{\frac{k}{m} }

    ==> \frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }  x  2\pi \sqrt{\frac{m}{k} }

    ==> 1/3.742 = 0.27

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