A small block of mass 1 kg is placed on a horizontal table, connected by a weightless spring to the central rod of the table as shown in the adjoining figure.
the table is set to rotate at a speed of 200 rpm which causes spring to stretch by 5 cm. If the original length of the spring is 30 cm, determine its force constant.

Explanation

Mass of block, m = 1kg
frequency of rotation of table,v = 200/60 = 3.33 rps
Extension produced in the spring, x = 5 cm
Length of stretched spring, r = 30 + x = 30 + 5 = 35 cm
When the table is set into rotation, the tension in the spring is equal to the centripetal force, So,
Restoring force an spring = Centripetal force
`rArr ky = mromega^(2) = mr (2piv)^(2)`
`rArr k = (4pi^(2)v^(2)mr)/(y)`
` = (4 xx 9.87 xx (3.33)^(2) xx 1 xx 35 xx 10^(-2))/(5 xx 10^(-2)) = 3065` N/m