1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spring and Elevator Questions i need help am stuck

  1. Aug 16, 2008 #1
    A particle of mass 0.2kg hangs from and ideal spring. In equilibrium, the spring is stretched by amount of 0.49cm. The spring is then suspended from the ceiling of a lift and hangs motionless relative to the lift as the lift descends with a constant velocity of 2.0m/s. The lift then suddenly stops.

    a) With what amplitude will the particle oscillate?
    b) What is the equation of the motion for the particles?

    Take the upward directions to be positive and g= 9.8ms-2

    solution for question a)

    I know that at the point of the elevator stops. Vmax= 2.0m/s
    at equilibirium
    0.2*9.8= -(k)(0.0049)
    w=-sqrt(400/0.2)= -44.72

    using v(t)= -wA sin(wt + teta)
    at equilibirium teta= 0, v(t)=Vmax

    2=-44.72*A sin (wt)

    A= -(2/(44.72*sin(wt))

    I am stuck now...suddenly my brain jammed....what is the value of the t here?
    t= 0 or t= 2?

    or since it is at equilibrium when at the point release. therefore

    Vm= wA
    Which then,
    A= -(2/44.72)= 0.044722719

    Please help...
  2. jcsd
  3. Aug 16, 2008 #2
    I believe that the second part you wrote is correct. 4.4 cm seems like a reasonable amplitude for this problem, as well, since the spring is strong but the elevator was moving relatively fast.

    The first way you tried would also work. The block/spring system is still in equilibrium when the elevator stops, so you can call that t=0. But its first maximum amplitude doesn't occur until the sin() term is at a maximum, which means wt = pi/2 and your equation simplifies to what you wrote later since sin(pi/2) is one. You don't really need to find t, but you could if you wanted.
  4. Aug 16, 2008 #3
    Thanks..i have to agree with you on the that hight amplitude for sin(pi/2) whereas cos (0)

    The answer should be correct in my rational thinking...:)

    thank you again...
  5. Aug 17, 2008 #4
    for the second part of the questions...
    b) What is the equation of the motion for the particles?

    "equation of the motion for the particles" in another word...the equation for the displacment for the particles..since the particle is oscillating.

    Am i right to say i will be using the displacement equations.

    x(t)= A cos (wt + teta)

    i konw that (wt + teta) is the initial phase of the particle oscillating...but i dun think is right for me to put pi/2 as the teta...i believe that this is right coz...i know that the value of teta depends on the displacement and the velocity at t=0. therefor that teta is fixed constant value...

    putting the all the values in will get

    x(t)= 0.044 cos (44.721 t + pi/2)

    should i put back the minus sign??

    am i right or wrong here?

    is this the equation?

    i stuck with my thoughts here...???
    Last edited: Aug 17, 2008
  6. Aug 17, 2008 #5
    What you wrote is correct, I think, but you can also re-write the expression with a sin() term, since [tex]cos{\left(\theta + \pi/2\right)} = sin{\theta}[/tex].

    Some people like using sin() because it shows more clearly that the position is 0 at t=0. cos() is used often because most of the time when learning SHM you are talking about a spring which is displaced and then released from rest, which is not the case here.

    Whether you put in a minus sign or not depends on which direction (up/down) you are calling positive.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Spring and Elevator Questions i need help am stuck