A mass on a spring with constant 2.76N/m vibrates with position given by the equ

Question

A mass on a spring with constant 2.76N/m vibrates with position given by the equation x = (5.00cm) cos(3.80t rad/s). (a) During the first cycle, for 0 <t < 1.65 s, just when isthe system's potential energy of the system changing most rapidlyinto kinetic energy?
first time 1 s second time 2 s
(b) What is the maximum rate of energy transformation?
3 mW (a) During the first cycle, for 0 <t < 1.65 s, just when isthe system's potential energy of the system changing most rapidlyinto kinetic energy?
first time 1 s second time 2 s
(b) What is the maximum rate of energy transformation?
3 mW first time 1 s second time 2 s

Explanation / Answer

   We havepotential energy U = 1/2kx^2             = (1/2)(2.76N/m)(0.05mcos3.8trad/s)^2             = 0.00345cos23.8t    Now we need to findthe time where the change in the potential energy is maximum    Now dU/dt =0.00345*2cos3.8t*-sin3.8t       = -0.00345sin7.6t Now the change is maximum when sin7.6t is 1    therefore 7.6t= /2 or 7.6t = 5/2    or t = 0.206sand t = 1.03s
Maximum rate of energy tranfer is 0.00345W
Hope this helps you

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