A body of mass 4.9 ~kg hangs from a spring and oscillates with a period 0.5 ~s . On the removal of the body, the spring is shortened by (take, g=10 ~ms ⁻², ^2=10 )

Step-by-step solution

Given that, mass of body, m=4.9 ~kg Time period of oscillation of spring, T=0.5 ~s Acceleration due to gravity, g=10 ~m / s ^2 and ^2=10 We know that, time period of spring T=2 m k By squaring on both sides, T^2=4 ^2 m k m k = T^2 4 ^2 = (0.5)^2 4 10 = 0.25 40 =0.00625 After removal of mass, length of spring decrease is equal to extension produced in spring. By using equilibrium condition, F=m g k x=m gx= m k g Substituting the values, we get aligned x & =0.00625 10 & =0.0625 ~m & =6.25 ~cm aligned Hence, the sprin