A 0.84 kg mass is attached to a light spring with a force constant of 42.9 N/m a

Question

A 0.84 kg mass is attached to a light spring with a force constant of 42.9 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

a)

½ * m* v^2 = PEmax = ½ * k* A^2

=> A* sqrt(k/m) = 0.05* sqrt(42.9/0.84) = 0.357 m/s

b)

total E = max U = ½kA² = ½ * 42.9N/m * (0.05m)² = 0.0536 J

At x = 1.5 cm, U = ½ * 42.9N/m * (0.015m)² = 0.00483 J

leaving KE = 0.0536J - 0.00483J = 0.04877 J = ½mv² = ½ * 0.84kg * v²

v = 0.34 m/s

c)

That's the same as part b, no?

d)

If the speed is ½ of the maximum value, then the KE is ¼ of the maximum value, and the spring energy is 3/4 of the maximum value. So

U = (3/4) * 0.0536J = 0.0402 J = ½ * 42.9N/m * x²

x = 0.04329 m

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