1) a) What is themechanical energy E of thelinear oscillator of Problem 1 above.

Question

1) a) What is themechanical energy E of thelinear oscillator of Problem 1 above. (Initially,

the block’sposition is x= 11 cm and its speed isv = 0. Spring constant k is 65N/m.)

b) What are thepotential energy and kinetic energy of the oscillator when theblock is at

x= A/2?(A represents the amplitude ofoscillations)

2)A performer seated on a trapeze is swingingback and forth with a period of 8.85 s. If she

stands up, thus raising the center of mass of thetrapeze + performer system by 35.0 cm,

what will be the new period of the system?(Ignore air friction and treat this as a small

angle oscillation)

Answer: 8.77 s

Explanation / Answer

1. x = A = 11 cm    ( since the speed v = 0 )     E = (1/2)kA2 = (1/2)65*11E-22 = 0.39 J     x = A/2 = 5.5 cm     Pot E = (1/2)kx2 = (1/2)65*5.5E-22 = 0.098 J     Kinetic E = (1/2)kA2 -(1/2)kx2 = (1/2)kA2 - (1/2)k(A2/4)= (3/8)kA2 = 0.295 J 2. T = 2(l/g)     l = T2g/42 = 19.44m After she stands up:     l'= l - l = 19.44 - 0.35 = 19.09 m The new period:    T' = 2(l'/g) = 2(19.09/9.8)= 8.77 s

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