AP EAPCET 24-Aug-2021 Shift 2 Paper

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 π2=10
We know that, time period of spring
T=2π√

m

k

By squaring on both sides, T2=4π2‌

m

k

⇒

m

k

=

T2

4π2

=

(0.5)2

4×10

=

0.25

40

=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=

m

k

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.