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

Answer

`1cm` .

Explanation

Here, `k = 100 N//m, l_(0) = 0.5 m, m =0.5kg`
`omega = 2 rad//s`
Let elongation of the spring be l
`F = kl` which provides the necessary centripetal
force `m omega^(2) r = m omega^(2) (l + l_(0))`
`:. Kl = m omega^(2) (l + l_(0))`
`l = (m omega^(2) l_(0))/( k - m omega^(2)) = (0.5 xx 2^(2) xx 0.5)/(100 - 0.5 xx 4) = (1)/(98)m = 1cm` .