What is the velocity at t = 4.0
seconds?
v(t) = íAȦ sin (Ȧt + ij)
v(4.0 s) = (í 2 ʌ) sin (8ʌ/3) m/s
v(4.0 s) = (í6.28)(0.866) m/s
v(4.0 s) = í5.4 m/s
14.8 - Acceleration
For SHM to occur, the net force on an object has to be proportional and opposite in sign
to its displacement. Again, we use the example of a mass attached to a spring on a
friction-free surface, like an air hockey table.
With a spring like the one shown in Concept 1 to the right, Hooke’s law (F = íkx)
states the relationship between net force and displacement from equilibrium. This
equation for force enables you to determine where the acceleration is the greatest, and
where it equals zero. The magnitudes of the force and the acceleration are greatest at
the extremes of the motion, where x itself is the greatest. This is the point where the
object is changing direction. Conversely, x = 0 at the equilibrium point, so F = 0 and
the object is not accelerating there.
The first equation shown in Equation 1 enables you to calculate the acceleration of an
object in SHM as a function of time. This equation can be simplified by noting that the
amplitude times the cosine function, the rightmost term in the equation, is the function
for the object’s displacement, x(t). We replace the terms A cos Ȧt by x(t) to derive the
second equation, which relates the acceleration directly to the object’s displacement.
This equation says that the acceleration at a particular time equals the negative of the
angular frequency squared times the object’s displacement at that time.
Finally, the third equation reveals that the maximum acceleration of the object is the amplitude times the square of the angular frequency. This
equation is a consequence of the first equation.Acceleration in SHM
Proportional to force
·Zero at equilibrium
·Maximum at extremes(^282) Copyright Kinetic Books Co. 2000-2007 Chapter 14