A mass of 148 g stretches a spring 13 cm. The mass is set in motion from its equlibrium position with a downward velocity of 10 cm/s and no

Question

A mass of 148 g stretches a spring 13 cm. The mass is set in motion from its equlibrium position with a downward velocity of 10 cm/s and no damping is applied.
(a) Determine the position u of the mass at any time t. Use 9.8 m/s as the acceleration due to gravity. Pay close attention to the units.
(b) When does the mass first return to its equilibrium position?

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Sigridomena 2 months 2021-07-31T10:38:16+00:00 1 Answers 2 views 0

Answers ( )

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    2021-07-31T10:39:43+00:00

    Answer:

    u(t)=1.15 \sin (8.68t)cm

    0.3619sec

    Explanation:

    Given that

    Mass,m=148 g

    Length,L=13 cm

    Velocity,u'(0)=10 cm/s

    We have to find the position u of the mass at any time t

    We know that

    \omega_0=\sqrt{\frac{g}{L}}\\\\=\sqrt{\frac{980}{13}}\\\\=8.68 rad/s

    Where g=980 cm/s^2

    u(t)=Acos8.68 t+Bsin 8.68t

    u(0)=0

    Substitute the value

    A=0\\u'(t)=-8.68Asin8.68t+8.68 Bcos8.86 t

    Substitute u'(0)=10

    8.68B=10

    B=\frac{10}{8.68}=1.15

    Substitute the values

    u(t)=1.15 \sin (8.68t)cm

    Period =T = 2π/8.68

    After half period

    π/8.68 it returns to equilibruim

    π/8.68 = 0.3619sec

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