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: Simple harmonic motion of a spring and mass

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data
    A mass hanging from a spring is displaced and released so that it vibrates vertically. Its
    maximum height above a tabletop is 30 cm and its minimum height above the table top
    is 12 cm. The mass vibrates 20 times per minute. At time 0, its height is 30 cm.

    Calculate the maximum velocity of the mass, and the times at which it occurs

    2. Relevant equations
    s(t) represents displacement
    v(t) represents s'(t) which is velocity

    3. The attempt at a solution
    Since we start at a maximum, we must use a cosine function
    the period T is 1/20 min. [itex] k = 2\pi / T = 40\pi [/itex]
    amplitude = (max - min) / 2 = [itex] \frac{30-12}{2} = 9 cm [/itex]
    [tex] s(t) = 9\cos \left(40\pi t\right) + 21 [/tex]
    [tex] v(t) = -360\pi \sin\left(40\pi t\right) [/tex]
    [tex] a(t) = -1440 \pi^2 \cos\left(40\pi t\right) [/tex]

    we find the max velocity when we solve a(t) = 0
    so we need to solve
    [tex] cos \left( 40\pi t \right) = 0 [/tex]

    since cos is negative for all half pi values, and consider only one rotation, for now...

    [tex] 40 \pi t = \frac{\pi}{2} \text{or} \frac{3\pi}{2}[/tex]

    which yields [tex] t = \frac{1}{80} \text{or} \frac{3}{80} [/tex]
    both answers have units of minutes.

    Since the function v(t) first decreases and then increases, the critical point 3/80 is where the max occurs
    if we plug each into v(t) we get
    [tex] v \left(\frac{3}{80}\right) = 1131 cm/min

    Is this all correct? Thank you for your help!
  2. jcsd
  3. Apr 30, 2013 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    It all looks correct to me.

    I would have been inclined to express the time in seconds rather than minutes. T = 3 sec.

    As far as max velocity is concerned:

    You know that the output of the sine function goes from -1 to 1, inclusive so if ##\displaystyle \ v(t) = -360\pi \sin\left(40\pi t\right)\,,\ ## then vmax = 360π . This gives an additional check regarding the fact the you have the correct time for max velocity.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted