Question

A 0.24 kg mass is attached to a light spring with a force constant of 30.9 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

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

Answer 1

a.

½ * k * x² = ½ * m* v²

½ * 30.9N/m * (0.05m)² = ½ * 0.24 * v²

=> v = 0.567 m/s

b. speed of the oscillating mass when the spring is compressed 1.5 cm"

total E = max U = ½kA² = ½ * 30.9N/m * (0.05m)² = 0.0386 J

At x = 1.5 cm, U = ½ * 30.9N/m * (0.015m)² = 0.00348 J

leaving KE = 0.0386J - 0.00348J = 0.0351 J = ½mv² = ½ * 0.24kg * v²

v = 0.541 m/s (= 54.1 cm/s )

"c. speed of the oscillating mass as it passes the point 1.5 cm from the equilibrium position"

That's the same as part b, = 0.541 m/s

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

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) * 0.0386J = 0.02895 J = ½ * 30.9N/m * x²

x = 0.0433 m (= 4.33 cm)