One end of a massless spring of spring constant 100 N/m and natural 0.5 m is fixed and the other end is connected to a particle of mass 0.5 kg lying on a frictionless horizontal table. The spring remains horizontal. If the mass is made to rotate at an angular velocity of 2 rad/s, find the elongation of the spring.

One end of a massless spring of spring constant 100 N/m and natural 0.

1.	One end of a massless spring of spring constant 100 N/m and natural 0.5 m is fixed and the other end is connected to a particle of mass 0.5 kg lying on a frictionless horizontal table. The spring remains horizontal. If the mass is made to rotate at an angular velocity of 2 rad/s, find the elongation of the spring.
Answer» Solution :The restoring FORCE in the spring provides the centripetal force to the mass. If .x. is the EXTENSION in the spring, then `Kx=m(L+x)omega^(2)` Putting the VALUES, we get `x=(0.5xx4xx0.5)/(100-0.5xx4)` `x=1 CM`

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