A 0.1 kg block is attached to a spring with a constant 2.5 N/mand oscillates on
ID: 1674514 • Letter: A
Question
A 0.1 kg block is attached to a spring with a constant 2.5 N/mand oscillates on a frictionless table. It's velocity is 0.2m/swhen x=0.05. What is the amplitude of oscillations? What is the ball's maximum acceleration? What is the ball's position when the acceleration ismaximum? What is the speed of the ball when x=0.03m? A 0.1 kg block is attached to a spring with a constant 2.5 N/mand oscillates on a frictionless table. It's velocity is 0.2m/swhen x=0.05. What is the amplitude of oscillations? What is the ball's maximum acceleration? What is the ball's position when the acceleration ismaximum? What is the speed of the ball when x=0.03m?Explanation / Answer
The equation of motion of the block attached to the springis x = Acos ---------(1) or (dx/dt) = -Awsin or v = -Awsin -----------(2) The angular frequency of oscillations is w = (k/m)1/2 where k = 2.5 N/m and m = 0.1 kg or w = (2.5/0.1)1/2 = 5 rad/s From equation (1) we get A = (x/cos) ------------(3) From equations (2) and (3) we get v = -(x/cos)wsin = -wx * tan------------(4) or = tan-1(-v/wx) = tan-1(-0.2/5* 0.05) = tan-1(-0.8) = -38.6o From equation (3) the amplitude of oscillations is A = (0.05/cos(-38.6o)) = 0.064 m The ball's maximum acceleration is amax = w2A or amax = (5)2 * 0.064 = 25 * 0.064 =1.6 m/s2 The position of the ball when the acceleration is maximum isequal to the amplitude of the ball that is A = 0.064 m from theequilibrium position. To find the speed of the ball when x = 0.03 m we first findthe angle made by the block from equation (1) that is = cos-1(x/A) ------------(5) Then we find the speed of the ball from equation (4) thatis v = -wx * tan The value of is obtained from equation (5)Related Questions
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