A block of mass 0.5 kg is attached to one end of a spring of spring constant 200 N m ⁻¹ , while the other end of the spring is fixed to a point on a smooth horizontal table. The block is made to move in a horizontal circle about the fixed point. If the extension in the spring during this motion is one-third of its natural length, the constant angular speed of the block is:

Step-by-step solution

Let the natural length of the spring be l . The extension in the spring is given as x = l 3 . The radius of the circular path is the total length of the stretched spring, R = l + x = l + l 3 = 4l 3 . The centripetal force required for circular motion is provided by the spring force: kx = mR ^2 Substituting the values: k ( l 3 ) = m ( 4l 3 ) ^2 Cancelling l/3 from both sides: k = 4m ^2 Given k = 200 N m ⁻¹ and m = 0.5 kg : 200 = 4(0.5) ^2 200 = 2 ^2 ^2 = 100 = 10 rad s ⁻¹ Answer: 10 rad s ⁻¹