One end of a spring of negligible unstretched length and spring constant k is fixed at the origin (0,0). A point particle of mass m carrying a positive charge q is attached at its other end. The entire system is kept on a smooth horizontal surface. When a point dipole p→ pointing towards the charge q is fixed at the origin, the spring gets stretched to a length ℓ and attains a new equilibrium position (see figure below). If the point mass is now displaced slightly by Δℓ≪ℓ from its equilibrium position and released, it is found to oscillate at frequency1δkm. The value of δ is ______
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
answer is 3.14.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Δℓ→x At ℓ:Fe=FSPkℓ=2kpqℓ3Fnet =Fsp−Fe=k(ℓ+x)−q(2kp)(ℓ+x)3=k(x+ℓ)−q(2kp)ℓ3(1+x/ℓ)3=kx+kℓ−q2kpℓ31−3xℓ=kx+kℓ−q2kpℓ3+2kpqℓ3⋅3xℓ⇒Fnet=kx+kℓ3xℓ=4kxkeq=4k ∴ T=2πm4k=πmk⇒f=1πkm So, δ=π=3.14