Q.

A body executes simple harmonic motion. The potential energy (PE), kinetic energy (KE) and total energy (TE) are measured as a function of displacement x. Which of the following statement is true?

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a

TE is zero when x = 0

b

PE is maximum when x = 0

c

KE is maximum when x = 0

d

KE is maximum when x is maximum

answer is C.

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Detailed Solution

y = xPE = 12mω2y2KE = 12mω2(a2-y2)TE = PE + KE       = 12mω2a2Since PE is maximum at x = a and KE is maximum at x = 0, therefore TE remains constant throughout the motion.
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