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Q.

A spring mass system ( mass m , spring constant k and natural length l ) rests in equilibrium on a horizontal disc. The free end of the spring is fixed at the center of the disc. If the disc together with spring mass system, rotates about it’s axis with an angular velocity ω,k>>mω2 the relative change in the length of the spring is best given by the option

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a

2mω2k

b

mω2k

c

23mω2k

d

mω23k

answer is B.

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Detailed Solution

Spring force=centrifugal force  kx=ml+xω2⇒kx−mxω2=mlω2     ⇒x=mlω2k−mω2      ⇒x=mlω2k      [k>>mω2]       Relative change is  xl=mω2k
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A spring mass system ( mass m , spring constant k and natural length l ) rests in equilibrium on a horizontal disc. The free end of the spring is fixed at the center of the disc. If the disc together with spring mass system, rotates about it’s axis with an angular velocity ω,k>>mω2 the relative change in the length of the spring is best given by the option