A 2.00kg object is attached to a spring. The force constant of the spring is 20.

Question

A 2.00kg object is attached to a spring. The force constant of the spring is 20.0 N/m. The object is displaced 0.0500m from the equilibrium position. After release, the object exhibits SHM. Determine the following:

            A) Period of the motion: ANS (2.81 s)

            B) Total energy of the system (0.0250 J)

            C) Object’s speed when it is 0.0100 meter from equilibrium (0.154 m/s)

Homework Physics 1 College level.

Please show all work and diagram as I already have the answers Thanks!

Explanation / Answer

A)We know that, time period in SHM is given by

T = 2 pi sqrt(m/k)

T = 2 x 3.14 x sqrt(2/20) = 1.99 s

Hence, T = 1.99 s

B)Total energy would be:

E = 1/2 k A^2

E = 0.5 x 20 x 0.05^2 = 0.025 J

Hence, E = 0.025 J

C)speed at 0.01 from equilibrum will be:

v = (2 pi/T)sqrt [A^2 - x^2]

v = [2 x 3.14/1.99] sqrt (0.05^2 - 0.01^2) = 0.154 m/s

Hence, v = 0.154 m/s

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