http://www.ck12.org Chapter 10. Periodic Motion
Notice that when the object is in the first quadrant, thex−component of the acceleration−axis negative whereas
thex−component of the displacement+xis positive and that when the object is in the second quadrant, the
x−component of the acceleration is positive+axand thex−component of the displacement−xis negative.
SinceF=ma,Fxandaxmust be in the same direction, and thereforeFxis oppositely directed toxas we have insisted
for SHM. We have also shown that the acceleration for uniform circular motion can be expressed asa=^4 π
2
T^2 r. But
we can write the accelerationaxasax=acosθand rearranging writea=cosaxθ. Substitutingcosaxθforaina=^4 π
2
T^2 r
yieldsax=^4 π
2
T^2 rcosθ. Sincercosθ=xsubstitution yields,ax=
4 π^2
T^2 x. Butaxandxare oppositely directed thus
ax=^4 π
2
T^2 x→ax=
4 π^2
T^2 (−x)(Equation 1). The quantity4 π^2
T^2 is a constant and soais directly proportional to the
displacement−x. But the forceFis directly proportional to the accelerationaas well sinceF=ma. Therefore, the
force is directly proportional to the opposite of the displacement.
We have shown two conditions for uniform circular motion: (1) the force and displacement are oppositely directed
and (2) the force and displacement are directly proportional. These are the same conditions for SHM.