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