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calculus

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Question 1:
Experimentally it is found that a 6 kg weight stretches a certain spring 6 cm. If the weight is pulled 4 cm below the equilibrium position and released:
a. Set up the differential equation and associated conditions describing the motion.
b. Find the position of the weight as a function of time.
c. Find the amplitude, period, and frequency of motion
d. Determine the position, velocity and acceleration of the weight 0.5s after it has been released.

Question 2:
A 0.25 kg mass is horizontally attached to a spring with a stiffness 4 N/m. The damping constant b for the system is 1 N-sec/m. If the mass is displaced 0.5 m to the left and given an initial velocity of 1 m/sec to the left, find the equation of motion. What is the maximum displacement that the mass will attain?

  • calculus -

    Question 1:

    The standard differential equation of motion is:

    mx"+Bx'+kx=f(t) ....(1)
    where
    x" is the second derivative with respect to time of displacement, x
    x' is the first derivative
    m=mass
    B=damping
    k=spring constant
    f(t)=applied external force

    For the given case,
    m=6 kg
    B=0
    k=mg/h=6*9.81/0.06=981 N/m
    f(t)=0

    Initial conditions:
    x(0)=0.04 (note: we use the convention positive downwards)
    x'(0)=0

    So substitute the quantities to equation 1:
    6x"+981x=0 .... (1a)
    Auxiliary equation:
    6m²+981=0
    m=±12.79i => α=a=0, β=b=12.79
    So solution to equation (1a)
    x=C1 cos(bt)+C2 sin(bt)
    Apply initial conditions:
    x(0)=0.04 = C1(1)+C2(0) => C1=0.04
    x'(t)=-C1*b*sin(bt)+C2*b*cos(bt)
    x'(0)=0 = C1(0)+C2*b(1) => C2=0/b=0
    Therefore C1=0.04, C2=0, or
    x(t)=0.04cos(12.79t)
    x'(t) = d(x(t))/dt = -12.79*0.04sin(12.79t)=-0.511sin(12.79t)


    I will leave it to you to answer parts (c) and (d).


    Q2 has been answered previously, see:
    http://www.jiskha.com/display.cgi?id=1299811520

  • calculus -

    thanxxxxx

  • calculus -

    thanxxxxx...what is * stand for??

  • calculus -

    If f(x) is a function, then f'(x)=dy/dx.

    If the independent variable is known or understood (often either x or t), then
    y'=dy/dx or dy/dt, depending on the context.

    In this case, the independent variable understood is t, so x(t) is the function, x' stands for x'(t)=dx/dt, and x" stands for x"(t)=d²x/dt².

  • calculus -

    i still don't get it the answer for question 1, c and d..can u give me the answer please..

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