Asked by jake
One end of a spring is attached to a wall and the other end is attached to a 20.4 gram block resting on a horizontal frictionless surface. The spring constant for the spring is 138 N/m. The block is moved parallel to the spring axis in such a way that its speed is 8.80 m/s when it passes through the unstrained spring position. What is the amplitude of the resulting simple harmonic motion?
Answers
Answered by
Anonymous
if F = -k x = m a
and x = A sin w t where w = 2pi f
then
v = dx/dt = A w cos w t
and
a = d^2x/dt^2 = - A w^2 sin w t = - w^2 x
then
-kx = - m w^2 x
and w = sqrt (k/m) but you know that, right?
so here
w = 2 pi f = sqrt ( 138 / 0.0204) = 82.2
now the max speed = A w cos 0 = A w = A* 82.2 = 8.80
so A = 8.80 / 82.2= 0.107 meters = 10.7 centimeters
and x = A sin w t where w = 2pi f
then
v = dx/dt = A w cos w t
and
a = d^2x/dt^2 = - A w^2 sin w t = - w^2 x
then
-kx = - m w^2 x
and w = sqrt (k/m) but you know that, right?
so here
w = 2 pi f = sqrt ( 138 / 0.0204) = 82.2
now the max speed = A w cos 0 = A w = A* 82.2 = 8.80
so A = 8.80 / 82.2= 0.107 meters = 10.7 centimeters
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