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PE of a particle is . Then the time period of small oscillation.
detailed solution
Correct option is A
by question PE=U(x)=ax2−bx---(1) to find force F=dUdx hence differentiate eqn(1) wrt x F=-2ax3+bx2---(2) at mean postion F=0, apply to eqn(2) 2ax3=bx2 ⇒x=2ab---(3) to find force constant, K=d2Udx2x=2ab K=6ax4-2bx3x=2ab∵dFdx=ddx-2ax3+bx2=ddx-2ax-3+bx-2=6ax4-2bx3 K=6a2ab4-2b2ab3 K=b48a3 =ω2m; ω=angular velocity;m=mass;T=time period ω=b48a3m=2πT T=2π8a3mb4Talk to our academic expert!
Similar Questions
A body executes S.H.M. under the action of a force F1 with a time period 7/6 seconds. If the force is changed to F2 it executes S.H.M. with time period 7/8 seconds. If both the forces F1 and F2 act simultaneously in the same direction on the body, then its time period in seconds is
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