A 4.0 kg mass is attached to a spring that has a springconstant, k = 2500 N/m, a

Question

   A 4.0 kg mass is attached to a spring that has a springconstant, k = 2500 N/m, and is resting on a frictionless horizontalsurface. The spring is stretched by pulling on the mass untilthe amount of potential energy in the spring is 4.5 J. Theblock is then released and oscillates in simple harmonicmotion.

a.       (5 points) What is theamplitude of the oscillation?

b.     (5 points) What is the maximum velocity of the block?

c.       (5 points) How longdoes it take for the block to go from maximum spring length tominimum spring length?

d.     (5 points) What is the velocity of the mass when it is 1.5 cmfrom the spring’s equilibrium point?

Please show all Formulas and WorkUsed to get the answer thank you

Explanation / Answer

given that mass = 4.0kg spring constant = k = 2500N/m potential energy = 4.5J potential energy = 1/2kx2 = 4.5J we know that = k/m = 2500/4 =25rad/s total eneergy = E = 4.5J amplitude = A = 2E/k = 2*4.5/2500 =0.06m vmax = A = 25 * 0.06 = 1.5m/s when x = 1.5cm = 0.015m v = ±k/m (A2 - x2) =------m/s time taken to maxium to minum is T/2 but T = 2/ = -----s T/2 = -------s

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