Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

SHM - oscillating mass

  1. May 3, 2010 #1
    Would somebody kindly check I have approached this problem correctly, please?
    Thank you.

    1. The problem statement, all variables and given/known data

    A 0.2kg mass is suspended from a light spring, producing an extension of 5.0cm.

    a) Calculate the force constant of the spring.

    The mass is now pulled down a further 2.0cm and released.

    b) Calculate
    i) the time period of the resultant oscillations
    ii) the maximum velocity of the mass

    c) sketch a displacement time graph for one complete oscillation of the mass, marking amplitude (A) and time period (T) and showing where maximum velocity occurs

    2. Relevant equations

    F = k\Delta x \\
    T = 2\pi \sqrt {\frac{m}{k}} \\
    f = \frac{1}{T} \\
    V_{\max } = 2\pi fA \\

    3. The attempt at a solution

    a){\rm{ }}k = \frac{{20}}{{0.05}} \\
    = 400 \\
    b){\rm{ i) T = 2}}\pi \sqrt {\frac{2}{{400}}} \\
    = 0.4442s{\rm{ }}\left( {4s.f.} \right) \\
    {\rm{ii) f = }}\frac{1}{{0.4442}} \\
    = 2.2624Hz{\rm{ }}\left( {4s.f.} \right) \\
    V_{\max } = 2\pi \times 2.2624 \times 0.07 \\
    = 0.9951{\rm{ }}\left( {4s.f.} \right) \\
    1.0ms^{ - 1} {\rm{ }}\left( {2s.f.} \right) \\

    I have taken the Amplitude from the spring equilibrium and not the equilibrium when the mass attached to the spring with initial extension of 5.0cm. Is this correct or should the Amplitude be just the secondary extension of 2.0cm?
    Thank you
    Last edited: May 3, 2010
  2. jcsd
  3. May 3, 2010 #2


    User Avatar
    Homework Helper

    First, you need units!

    Part b) isn't correct. Why did you use 2 kg for m?

    Also, you need to use 2 cm, not 7 cm. This is because amplitude is always the displacement from the equilibrium point, and when the mass is attached, the equilibrium point is 2 cm away.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook