A 150 ball attached to a spring with spring constant 2.30 oscillates horizontall

Question

A 150 ball attached to a spring with spring constant 2.30 oscillates horizontally on a frictionless table. Its velocity is 20.0 when x= -5. What is the amplitude of oscillation? What is the speed of the ball when ?

Explanation / Answer

the above approach is correct > you may try this as well the ball is doing SHM x = a sin wt a = amplitude = ? k = 2.4 N/m, m = 0.150 kg w =angular freq = [k/m]^1/2 = 4 ---------------------------- velocity v = dx/dt = a w cos wt v^2 = (aw)^2 cos^2 wt v^2 = (aw)^2 [ 1- sin^2 wt] v^2 = (aw)^2 [ 1- (x/a)^2] v^2 = w^2 [a^2 - x^2] ------ (1) standard equ >> may remember a^2 = [v/w]^2 + x^2 ----------------------------------- when x = - 0.05, v = 0.20 m/s a^2 = [0.20/4]^2 + (0.05)^2 = (0.05)^2+ (0.05)^2 = 2(0.05)^2 a = 0.05 root 2 a = 0.0707 meter a = 7.07 cm >>>>>>>>>> your answer --------------------------------- x = 0.03 m v^2 = w^2 [a^2 - x^2] = 16[(2*(0.05)^2 - (0.03)^2] v^2 = 16[0.005 - 0.0009] = 0.0656 v = 0.256 m/s = 25.6 m/s >>>> answer ====================== remember > max speed = aw = 7.07*4 = 28.28 cm/s at x =0

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