A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium position with a downward velocity of 10 cm/s, and

Question

A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium position with a downward velocity of 10 cm/s, and if there is no damping, determine the position u of the mass at any time t. When does the mass first return to its equilibrium position

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Thu Hương 5 months 2021-08-17T11:57:09+00:00 1 Answers 0 views 0

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    2021-08-17T11:59:04+00:00

    Answer:

    w_{0}=14

    t=\frac{\pi }{14}

    Explanation:

    Data

    mass m= 100g

    Length L= 5cm

    we can use:

    gm-kL= 0

    divide both side by m

    g – \frac{kL}{m}=0

    where

    \frac{k}{m} = \frac{g}{L}

    \frac{k}{m}=w_{0}^{2}

    so now

    w_{0}^{2} = \frac{9.8*100}{5}

    w_{0}^{2}=\frac{980}{5}

    w_{0}^{2}=196

    square both side

    w_{0}=\sqrt{196}

    w_{0}=14

    We can apply:

    u(t)=Acoswt +Bsinwt

    u(t)=Acos14t +Bsin14t

    u(0)=0  where A=0

    therefore

    u(0) = Bsin14t  

    u^{'}(0) = 10 ⇒ 10=14B ⇒ B=\frac{14}{10} B=\frac{5}{7}

    so now u(t)=\frac{5}{7}sin14t

    so t will be:

    t=\frac{\pi }{14}

    t=\frac{3.14}{14}

    t=0.22 seconds

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