1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Spring harmonic motion

  1. Apr 16, 2009 #1
    1. The problem statement, all variables and given/known data

    1. The St. Louis Arch has a height of 192m. Suppose a stunt woman of mass 60kg jumps off the top of the arch with an elastic band attached to her feet. She reaches the ground at zero speed. Find her kinetic energy K after 2 sec of the flight. Assume that the elastic band obeys Hooke's law, and neglect its length when relaxed

    2. A 0.12kg block is suspended from a spring. When a small stone of mass 30g is placed on the block, the spring stretches an additional 5cm. With the stone on the block, the spring oscillates with an amplitude of 12cm. (a) What is the frequency of the motion ? (b) How long does the block take to travel from its lowest point to its highest point ? (c) What is the net force of the stone when it is at a point of maximum upward displacement ?

    2. Relevant equations

    w= sqrt(k/m)

    3. The attempt at a solution

    1. I use ky=mg to calculate k and get k=60*9.81/192=3.066 N/m
    Then I calculate w=sqrt(k/m)=0.226 rad/s

    Then I use v=-A*w*sin(wt) to calculate v at t=2 and get -19m/s and substitute into K=1/2 mv^2 and got 10.8kJ. However, I am wrong. Can you guy suggest me a way to fix my solution ??

    2. I used the same formula as number 1: ky=mg to get k and then w=sqrt(k/m)=9.04 rad/s. Finally I use f=w/2pi to get f=1.44Hz.

    For part b, T=1/f=0.694 sec and for part c, F=mg+kx= 119 N

    I am wrong again.Any suggestion ??
  2. jcsd
  3. Apr 16, 2009 #2


    User Avatar
    Homework Helper

    Stop right there. When the jumper reaches the ground with V = 0, and 0 kinetic energy, she is not in static equilibrium. Hence you need to examine the energy relationship.

    At the top the total energy is m*g*h, so neglecting the relaxed length, and the practicality of trying to envision it, that means that the PE at the top must equal the PE of the elastic at the bottom. So ...

    m*g*h = 1/2* k*x2

    Or k = 2mg/h
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook