http://www.ck12.org Chapter 10. Periodic Motion
FIGURE 10.6
Examples of such a spring pendulum would include a bungee jumper or just a large spring, as shown in the video
below:
MEDIA
Click image to the left for use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/62067The Period for a Mass on a Spring in SHM
The amount of time it takes an object to repeat its motion is called theperiodof the oscillations, written asT.
Imagine stretching the spring with the attached mass ofFigure10.6 to some position+xand releasing it. The
period is how much time is required for the mass to once again arrive at position+x.
Since we have shown that the projection of uniform circular motion is SHM and that a mass vibrating on a spring is
in SHM, we can use the results of uniform circular motion to answer this question.
We know that the magnitude of the force on the mass isF=kxwhen the spring is stretched a distancex. But we
also know thatFx=max, and somax=kx(Equation 2). Thex-component of the acceleration, however, isax=^4 π
2
T^2 x
(Equation 1). Substituting foraxfrom Equation 1 into Equation 2, we havem^4 π
2
T^2 x=kx→m4 π^2
T^2 =k. Solving for
the period,T→m^4 π
2
T^2 =k→T
(^2) = 4 π 2 m
k→T=^2 π
√m
k.
Check Your Understanding
Assume the block inFigure10.6 has a mass of 1.50-kg and is attached to a spring with spring constant 25.0 N/m.
What is the periodTof the block?
Answer:We knowT= 2 π
√m
k. Substitution into this equation gives usT=^2 π
√ 1. 50 kg
25. 0 Nm =^1.^539 →^1.^54 s.
In the study of vibrational motion, it is useful to define two other quantities in addition to the periodT.