A 0.35-kg mass at the end of a spring oscillates 2.7 times per second with an am

ID: 1556375 • Letter: A

Question

A 0.35-kg mass at the end of a spring oscillates 2.7 times per second with an amplitude of 0.15 m.

Part A

Determine the magnitude of the velocity when it passes the equilibrium point.

Express your answer using two significant figures.

Part B

Determine the magnitude of the velocity when it is 0.11 m from equilibrium.

Express your answer using two significant figures.

m/s

Part C

Determine the total energy of the system.

Express your answer using two significant figures.

Part D

Determine the equation describing the motion of the mass, assuming that at t = 0, x was a maximum and that t in seconds.

Determine the equation describing the motion of the mass, assuming that at = 0, was a maximum and that in seconds.

Total Energy = (1/2)kA2 = (1/2)mw2A2 = 1.1332 J

at equilibrium position, x = 0

so, K.E = 1.1332 J = (1/2)mv2

A] therefore at x = 0,v = 2.545 m/s

B] Elastic Potential Energy at x = 0.11 m, is E = (1/2)mw2(0.11)2 = 0.6094 J

so, K.E = 1.1332 - 0.6094 = 0.5238 J

hence v = 1.73 m/s

C] Total Energy = 1.1332 J

D] At t = 0s, x was maximum. Thus, the function is a cosine function with amplitude A = 0.15m

so, the function is, x = 0.15cos(17t).

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