An object is undergoing simple harmonic motion with frequency f = 3.5 Hz and an amplitude of 0.15 m. At t = 0.00 s the object is at x = 0.00

Question

An object is undergoing simple harmonic motion with frequency f = 3.5 Hz and an amplitude of 0.15 m. At t = 0.00 s the object is at x = 0.00 m. How long does it take the object to go from x = 0.00 m to x = 4.00×10−2 m.

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Bình An 5 years 2021-08-29T03:07:42+00:00 1 Answers 35 views 0

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  1. Edana
    0
    2021-08-29T03:09:26+00:00
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    Answer:

    Time taken by the object is 0.012 s .

    Explanation:

    Given :

    Frequency , f = 3.5 Hz .

    Amplitude , A = 0.15 m .

    At time t = 0 , x = 0 m.

    Since , at time t = 0 , x = 0 m .

    Therefore , equation of displacement is :

    x=Asin(\omega t)     …equation 1.

    Here , \omega is angular frequency and is given by :

    \omega=2\pi f=22\ Hz.

    We need to find the time at which its displacement is , x = 4.00\times 10^{-2}\ m.

    Putting all these value in equation 1 we get ,

    4\times 10^{-2}=0.15 \times sin(22\times t) \\\\0.27=sin(22\times t)\\\\22\times t=sin^{-1}{0.27}\\\\t=\dfrac{sin^{-1}0.27}{22}\\\\t=0.012\ s .

    Hence , this is the required solution.

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