A 0.47-kg object connected to a light spring with a force constant of 21.2 N/m o
ID: 2148992 • Letter: A
Question
A 0.47-kg object connected to a light spring with a force constant of 21.2 N/m oscillates on a frictionless horizontal surface. The spring is compressed 4.0 cm and released from rest.(a) Determine the maximum speed of the object.
(b) Determine the speed of the object when the spring is compressed 1.5 cm.
(c) Determine the speed of the object as it passes the point 1.5 cm from the equilibrium position.
(d) For what value of x does the speed equal one-half the maximum speed?
Explanation / Answer
From SHM we have x = A*cos(?t) and v = -A*?*sin(?t) where ? = sqrt(k/m) = sqrt(21.2/0.47) = 6.72 rad/s and A = 4.0cm So vmax = A*? = 4.0cm*6.72 rad/s = 26.9cm/s b) Find t for x = 1.5 and then plug into v eqn 1.5 = 4.0*cos(6.72*t) => t = arccos(1.5/4.0)/6.72 = 0.177s Now speed is the magnitude of velcoity so we will use the + ans v = -26.9*sin(6.72*0.177) = 24.9cm/s c) Since you asked for speed and not velocity the ans here is the same as before 24.9cm/s d) Now find time for 1/2v 13.45 = 26.9*sin(6.72*t) so t = arcsin(0.5)/6.72 = 0.0779s so x = 4.0cm*cos(6.72*0.0779) = 3.46 cm
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