Questions
PE of a particle is . Then the time period of small oscillation.
a
2π8a3mb4
b
2π8b3mb4
c
2π8a4mb3
d
2π8b4ma2
detailed solution
Correct option is A
The equilibrium position is dudx=0Because oscillatory force at mean position is ‘o’⇒−2ax03+bx02=0 or x0=2abu(x)=ax02−bx0x=x0d2udx2x=x0=KK=6ax04−2bx03=6a2ab4−2b2ab3=b48a3 Thus, u(x)=Ux0+12b48a3y2 Comparing with 12mω2y2=12b48a3y2ω=b48a3m or T=2π8a3mb4Talk to our academic expert!
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