A 0.90 kg mass is attached to a light spring with a force constant of 24.9 N/m a

Question

A 0.90 kg mass is attached to a light spring with a force constant of 24.9 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 Correct: Your answer is correct. m/s

(b) speed of the oscillating mass when the spring is compressed 1.5 cm 1.84 Incorrect: Your answer is incorrect. Is energy conserved for this oscillating system? 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

Vmax = A*w = A*sqrt(k/m) = 0.263 m/s

b) from energy conservation

0.5*k*x^2 + 0.5*m*v^2 = 0.5*K*A^2

k*x^2 + m*v^2 = k*A^2

(24.9*0.015*0.015) + (0.9*v^2) = 24.9*0.05*0.05

v = 0.2508 m/s

c) x1 = 5 -1.5 = 3.5
0.5*k*x1^2 + 0.5*m*v^2 = 0.5*K*A^2

k*x1^2 + m*v^2 = k*A^2

(24.9*0.035*0.035) + (0.9*v^2) = 24.9*0.05*0.05

v = 0.1878 m/s

d) v = Vmax/2 = 0.263/2 = 0.1315

0.5*k*x^2 + 0.5*m*v^2 = 0.5*K*A^2

k*x^2 + m*v^2 = k*A^2

(24.9*x^2) + (0.9*0.1315*0.1315) = (24.9*0.05*0.05)

x = 0.0433 m

x = 4.33 cm

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