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Homework Help: Amplitude and Period

  1. Apr 20, 2008 #1
    1. The problem statement, all variables and given/known data
    A massless spring of spring constant k = 74 N/m is hanging from the ceiling. a 490 g mass is hooked onto the unstretched spring and allowed to drop.

    A) Find the amplitude
    B) Find the period of the resulting motion

    2. Relevant equations
    f = 1/T
    x(t) = Acos (omega) t
    omega = sqrt( k / m )
    x(t) = Acos(omega * t + phi )
    omega = sqrt ( g / L )
    T = 2*PI * sqrt ( L / g )

    3. The attempt at a solution

    I'm lost in this chapter and have no idea where to start, i was curious if someone could help step me through this problem, would help me a ton.

  2. jcsd
  3. Apr 20, 2008 #2
    for part b)

    omega = sqrt (k / m)

    freq = omega / (2PI)

    T = 1/freq

    i cant figure out how to find amplitude
  4. Apr 20, 2008 #3


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    Homework Helper

    Hi nightshade123,

    Try using conservation of energy at the highest and lowest points.
  5. Apr 21, 2008 #4
    at the highest and lowest points v = 0 and Potential Energy is at its max and kinetic energy is 0

    i know i have this eqns

    E = U + K

    U = 1/2 K A^2 (cos(omega t ))^2
    K = 1/2 K A^2 (sin(omega t ))^2
    E = 1/2 K A^2
  6. Apr 21, 2008 #5
    1/2kx + 1/2mv = 1/2kA

    kx + mv = kA

    v = 0

    kx = kA


    F = -k * x

    x = - F/k

    F = mg

    x = (mg)/k

    x = A
  7. Apr 21, 2008 #6
    The amplitude is the maximum extension in the spring.
    Calculate omega using omega = (k/m)^1/2.
    Now calculate extension using hook's law.
    Now for max extension = kx^2=mv^2
    Calculate velocity.Now use v= omega*Amplitude to get amplitude.
    Time period = 2pi/omega.
  8. Apr 21, 2008 #7
    you dont have to do all that work A = x work it out yourself. A = .065 m thx for the help tho
  9. Apr 21, 2008 #8


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    Homework Helper


    This part isn't quite right. In the formula for the energies the x, v, and A have to be squared. Also, you get x=A but it doesn't tell you what either of them is. The problem is in setting up the equation like this (where x=0 at the equilibrium point).

    Instead you could try setting up the equation with x=0 at the spring's unstretched point which would then explicitly include gravity in the energy equation. Letting x=0 and h=0 at the higest point, and setting enegy to be conserved between the higest and lowest points gives:

    \frac{1}{2}k x_i^2 + mgh_i &=\frac{1}{2}k x_f^2 + mgh_f\nonumber\\
    0+0 &= \frac{1}{2} k x^2 + m g (-x)\nonumber

    Solving this for the (nonzero) value of x would give the entire range of motion from highest to lowest point; the amplitude would be half of that.

    This alternative method looks good. The distance from the highest point to the equilibrium position is the amplitude and so the x value you found here equals A.
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