A block of mass 2.00 kg is suspended from an ideal spring with force constant k

Question

A block of mass 2.00 kg is suspended from an ideal spring with force constant k = 0.500 N/m. The mass is allowed to come to rest at its equilibrium position and then pulled down a distance of 0.800 m and pushed down with an initial velocity of -0.300 m/s. (Take the upward direction as a positive y-axis.) What is the period of the resulting SHM of the block? Show work. What is the amplitude ym of the motion? (Use conservation of energy.) The resulting motion of the block can be written as y = ym cos(omegat + phi) (y in m, t in s). Give the values of omega and phi. Show work.

Explanation / Answer

m = 2 kg
k = 0.5 N/m
Vmax = 0.3 m/s

y(0)=0.8 m

a) T=2pi*sqrt(m/k) =2pi*sqrt(2/0.5) =12.57 sec

b) Vmax = w* A

A = Vmax/w = 0.3/0.5= 0.6 m

c) w = sqr(0.5/2) = 0.5 rad/s

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