A mass weighting 48 lbs stretches a spring 6 inches. The mass is in a medium that exerts a viscous resistance of 27 lbs when the mass has a

Question

A mass weighting 48 lbs stretches a spring 6 inches. The mass is in a medium that exerts a viscous resistance of 27 lbs when the mass has a velocity of 6 ft/sec. Suppose the object is displaced an additional 6 inches and released.

Required:
a. Find an equation for the object’s displacement, u(t), in feet after t seconds.
b. What is the mass of the object?
c. What is the damping coefficient?
d. What is the spring constant?

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RI SƠ 3 years 2021-09-04T14:19:23+00:00 1 Answers 16 views 0

Answers ( )

  1. thienthanh
    0
    2021-09-04T14:20:36+00:00
    Reply

    Answer:

    a)

    u(t)=0.499ft.e^{-\frac{144.76lb/s}{2(48lb)}t}cos(\omega t)\\\\u(t)=0.499ft.e^{-1.5t}cos(\omega t)

    b)

    m = 48lb

    c)

    b = 144.76lb

    Explanation:

    The general equation of a damping oscillate motion is given by:

    u(t)=u_oe^{-\frac{b}{2m}t}cos(\omega t-\alpha)    (1)

    uo: initial position

    m: mass of the block

    b: damping coefficient

    w: angular frequency

    α: initial phase

    a. With the information given in the statement you replace the values of the parameters in (1). But first, you calculate the constant b by using the information about the viscous resistance force:

    |F_{vis}|=bv\\\\b=\frac{|F_{vis}|}{v}\\\\|F_{vis}|=27lbs=27*32.17ft.lb/s^2=868.59ft.lb/s^2\\\\b=\frac{868.59}{6}lb/s=144.76lb/s

    Then, you obtain by replacing in (1):

    6in = 0.499 ft

    u(t)=0.499ft.e^{-\frac{144.76lb/s}{2(48lb)}t}cos(\omega t)\\\\u(t)=0.499ft.e^{-1.5t}cos(\omega t)

    b.

    mass, m = 48lb

    c.

    b = 144.76 lb/s

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