A 0.66 kg mass is attached to a light spring with a force constant of 35.9 N/m a
ID: 1297350 • Letter: A
Question
A 0.66 kg mass is attached to a light spring with a force constant of 35.9 N/m and set into oscillation on a horizontal frictionless surface. If the spring is stretched 5.0 cm and released from rest, determine the following.
(a) maximum speed of the oscillating mass
??m/s
(b) speed of the oscillating mass when the spring is compressed 1.5 cm
??m/s
(c) speed of the oscillating mass as it passes the point 1.5 cm from the equilibrium position
??m/s
(d) value of x at which the speed of the oscillating mass is equal to one-half the maximum value
??m
Explanation / Answer
w = sqrt(K/m) = sqrt(35.9/0.66) = 7.38 rad/s
a)
Vmax = A*w = 0.05*7.38 = 0.369 m/s
#b)
TE = KE + PE
0.5*K*A^2 = 0.5*M*v^2 + 0.5*K*x^2
K*A^2 = M*v^2 + K*x^2
K*(A^2-x^2) = m*v^2
35.9*((0.05*0.05)-(0.015*0.015)) = 0.66*v^2
v = 0.352 m/s
#c)
v = 0.352 m/s
#d)
TE = KE + PE
0.5*K*A^2 = 0.5*M*v^2 + 0.5*K*x^2
K*A^2 = M*vmax/2^2 + K*x^2
K*(A^2-(x)^2) = m*vmax/2^2
35.9*((0.05*0.05)-(0.025*0.025)) = 0.66*0.1845*0.1845
x = 0.043 m = 4.3 cm
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