One end of a massless spring of 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 a frictionless horizontal table. The spring remains horizontal. If the mass is made to rotate at angular velocity of 2 rad / s , then elongation of spring is

Step-by-step solution

For the circular motion acceleration towards the centre is v^2 r . The horizontal force on the particle is due to the spring and k l , where l is the elongation and k is spring constant of spring. aligned & k l= m v^2 r & =m ^2 r=m ^2 (l₀+l ) & (k-m ^2 ) l & =m ^2 l₀ & l & = m ^2 l₀ k-m ^2 aligned Putting the values 1= 0.5 4 0.5 100-0.5 4 = 1 98 ~m =1 ~cm