A linear harmonic oscillator of force constant 2 × 106 N/m and amplitude 0.01m has a total mechanical energy of 160 joules. Its
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
Maximum potential energy is 100J
b
Maximum K.E. is 160J
c
Maximum P.E. is 160J
d
Minimum P.E. is zero
answer is C.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Elastic potential energy = 12kx2 == (2 × 106) (0.01)2 = 100 JThe energy is converted into kinetic energy∴maximum K.E. = 100 JGiven that total mechanical energy is 160 J , hence maximum potential energy is 160 J.