Given that, mass of body,
m=4.9‌kg Time period of oscillation of spring.
T=0.5s Acceleration due to gravity,
g=10m∕s2 and
Ï€2=10 We know that, time period of spring
T=2π√ By squaring on both sides,
T2=4π2‌ ⇒= ===000625 After removal of mass, length of spring decrease is equal to extension produced in spring.
By using equilibrium condition,
F=mg ⇒kx=mg x=g Substituting the values, we get
x=0.00625×10 =0.0625m =6.25‌cm Hence, the spring is shortened by
6.25‌cm after removal of mass.