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Q.

The total mechanical energy of a harmonic oscillator of amplitude 1 m and force constant 200 N/m is 100 J then

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a

The minimum PE is zero

b

the maximum PE is 100 J

c

the minimum PE is 50 J

d

the maximum PE is 50 J

answer is A.

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Detailed Solution

given force constant k=200N/mtotal energy TE=100 Jmaximum kinetic energy is KE  =12mw2A2         maximum kinetic energy KE=12kA2     maximum kinetic energy KE=12×200×12  ∴KEmax=100 JTotal energy TE=KEmax+PEmin100=100+PEminPEmin=0 J
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