The potential energy of a particle Ux executing SHM is given by
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
Ux=K2a−x2
b
Ux=K1x+K2x2+K3x3
c
Ux=Ae−bx
d
Ux=a constant
answer is A.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
PE of body in SHM at an instant PE=U=12mω2y2=12Ky2m is mass of oscillator, ω is angular velocity, y is displacement of oscillator, k is force constantIf the displacement , y=a−x then U=12Ka−x2=12Kx−a2