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Homework Help: Oscillations and Waves of a spring

  1. Jan 27, 2006 #1
    Can someone please help me with this question?:confused:

    A 2.00-kg object is attached to a spring and placed on a horizontal, smooth surface. A horizontal force of 20.0 N is required to hold the object at rest when it is pulled 0.200 m from its equilibrium position (the origin of the x axis) The object is now released from rest with an initial position of x= 0.200m, and it subsequenly undergoes simple harmonic oscillations. Find
    (a) the force constant of the spring,
    (b) the frequency of the oscillations, and
    (c) the maximum speed of the object. Where does this maximum speed occur?
    (d) Find the maximum acceleration of the object. Where does it occur?
    (e) Find the total energy of the oscillation system.
    (f) Find the speed
    (g) Find the acceleration of the object when its position is equal to one third of the maximum value.
  2. jcsd
  3. Jan 28, 2006 #2


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    Can you show some of your working/thoughts?
  4. Jan 28, 2006 #3
    Tina is that you? Please show us your work lol.
  5. Jan 28, 2006 #4
    my work

    Here is my work... I'm not 100% sure.. This questions is hard...
    Mass = 2.00 kg
    F= 20.0 N
    x = 0.200m

    (a) finding K
    F= 0.5kx^2 if that is the right formula
    20.0 N = 0.5k(0.2)^2
    K= 1000 N/m

    (b) Since w is also frequency but is in s^-1
    w= (k/m)^1/2
    w= (1000/2.00)^1/2
    w= 22.36 s^-1

    (c) Vmax = wA.... What is A? im treating A as being the same as x.. I dont know how to do this question.
    Vmax = (22.36)(0.2)
    Vmax = 4.472 m/s
    Where does it occur?? no idea what formula to use.

    (d) Amax = w^2A.. again?? is this the right formula?

    (e) Total energy?
    E = 0.5 kA^2 or is it kinetic energy + potential energy??

    (f) Find the speed.. formula for speed is
    v = w/k..
    so if this is right.. then
    22.36/1000.. so v is 0.02236

    (g) No idea

    Please Help me!:confused:
  6. Jan 29, 2006 #5


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    (a) The spring constant, [itex]k[/itex] is given by Hooke's law for springs: [tex]F = -kx [/tex]. I think you answer is wrong by a factor of ten.

    (b)It looks like you've worked out [itex]\omega[/itex] correctly but remember [tex]f = \frac{\omega}{2\pi}[/tex]. Your error in (a) follows through here.

    (c)Your formula is correct but there are follow through errors from the previous questions, A is amplitude or maximum displacement. As the spring is stretch it gains elastic potential energy. Once the spring is released and begins to recoil the elastic potential energy is converted to kinetic energy. When the spring is in neither compression or tension there is no elastic potential energy in the spring, the energy of the system must remain constant therefore all of the elastic potential energy must be in the form of kinetic energy. So where is the velocity the greatest?

    (d)Correct formula except [itex] a = -\omega^2 x [/itex] because the acceleration is always in the opposite direction to the displacement (towards the equilibrium position)

    (e)Your forumla [itex]E = \frac{1}{2} k A [/itex]is correct.

    (f)I don't understand the question. Find the speed when? The speed is variable in any mass/spring system.

    (g)[tex] a = -\omega^2 x [/tex] where x is displacement. You know the amplitude so [itex] x = \frac{1}{3}A [/itex]
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