A 0.54 kg mass is attached to a light spring with a force constant of 36.9 N/m a

Question

A 0.54 kg mass is attached to a light spring with a force constant of 36.9 N/m and set into oscillation on a horizontal frictionless surface. If the spring is stretched 5.0 cmand released from rest, determine the following.

(a) maximum speed of the oscillating mass

(I got A. It's .41 m/s)

(b) speed of the oscillating mass when the spring is compressed 1.5 cm

(c) speed of the oscillating mass as it passes the point 1.5 cm from the equilibrium position

(d) value of x at which the speed of the oscillating mass is equal to one-half the maximum value

Explanation / Answer

a) angular frequency w = sqrt(k/m)

w = sqrt(36.9 / 0.54) = 8.27

max speed = wA = 8.27 x 0.05 = 0.41 m/s

b) total energy = K.E. + P.E

kA^2 /2 = mv^2 /2 + kx^2 /2

36.9 x 0.05^2 = 0.54 x v^2 + 36.9 x 0.015^2

v = 0.39 m/s

c) v = 0.39 m/s

d) total energy = K.E. + P.E

kA^2 /2 = mv^2 /2 + kx^2 /2

36.9 x 0.05^2 = 0.54 x (0.41/2)^2 + 36.9x^2

x = 0.0434 m or 4.34 cm

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