At the highest point of block, spring is in its natural length. Hence, amplitude, A=
mg
k
Given, f=10Hz ω=2πf=20πrad∕s ∴ω=√
k
m
On squaring both sides, we get
k
m
=ω2 ⇒vmax=Aω =
m
k
gω=
gω
ω2
∴A=
m
ω
=
π2
20π
=
π
20
m∕s ∴A=
mg
k
=
g
ω2
=
π2
400π2
=
0.25
100
m=0.25cm Now, when extension is 0.20cm, then displacement from equilibrium position =0.25−0.20=0.05cm x=0.05cm ∴v=ω√A2−x2 =20π√(0.25)2−(0.05)2 =2π√6cm∕s =15.4cm∕s