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Question

# A particle of mass m is attached to a spring of natural length L and spring constant k. If it is rotated in a horizontal plane with an angular speed ω, then what will be the new length of the spring ?

A
kLkmω2
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B
mω2Lkmω2
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C
mω2Lk
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D
Lkmω2
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Solution

## The correct option is A kLk−mω2Spring will start to elongate, so that required centripetal force can be provided by the spring force. Let the elongation in the spring be x, so the radius of the particle for the circular path becomes r=L+x ∴Fspring=kx Applying the equation of circular dynamics along radial direction: mω2r=Fspring ...(i) ⇒ mω2(L+x)=kx ⇒mω2L+mω2x=kx ∴x=mω2Lk−mω2 New length of the spring is: r=L+x=kLk−mω2

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