3.(a) We know from class that the velocity of a particle is
given by the first derivative of the position vector,
i.e. v = dr/dt. For one dimensional motion,
this means that v = dx/dt. The one-dimensional
acceleration is given the second derivative of the
position, i.e. a = d2x/dt2 = dv/dt.
Using this information, we getv = dx/dt = (.25 m)(3.2
rad/s) cos[(3.2 rad/s)t + .13] = .8 m/s cos[(3.2 rad/s)t
+ .13]
a = dv/dt = (.8 m/s)(3.2 rad/s)(-1) sin[(3.2 rad/s)t +
.13] = -2.56 m/s2 sin[(3.2 rad/s)t + .13]
The maximum will occur when the sinusoidal part of the
solution has a magnitude of 1. Thus vmax = .8
m/s and amax = 2.56 m/s2.
(b) At t = 0 s, cos[(3.2 rad/s)t + .13] = cos(.13) =
.992, and sin[(3.2 rad/s)t + .13] = sin(.13) = .130. This
gives
v(0 s) = .794 m/s
a(0 s) = .333 m/s2
(c) For a spring system, we know that F = -kx. From
Newton's second law, we know that F = ma. Putting the two
together gives
-kx = ma
-k(.25 m sin[(3.2 rad/s)t + .13]) = m(-2.56 m/s2
sin[(3.2 rad/s)t + .13])
k = (5 kg)(2.56 m/s2)/(.25 m) = 51.2 N/m
4. The place to start on this problem is to draw the forces
the mass. If the mass is moved from its equilibrium
position, then the spring on the left will have a force
on it, given by F = -(55 N/m)x (This assumes that x=0 m
is the equilibrium position). The spring on the right
will have a force on the mass given by F = -(85 N/m)x.
Therefore, the total force on the mass is given byF =
-(55 N/m)x - (85 N/m)x = -(140 N/m)x.
Therefore, we just have the equation for a simple
spring system with a spring constant of 140 N/m. We know
that the solution to this equation is
x(t) = A cos(ot) + B sin(ot)
where o = (k/m)1/2
= 7.9 rad/s. In order to find A and B, we need to use the
initial conditions that x(0 s) = .05 m and v(0 s) = 0
(Remember v = dx/dt). Plugging in, we get
x(0 s) = .05 m = A
v(0 s) = 0 = B
Thus the formula for the position of the mass as a
function of time is x(t) = .05 m cos(7.9 rad/s)t.
