Question

A spring hangs ( unstretched) from a support. A 300 gram mass is suspended from the spring and allowed to reach equilibrium. If the mass is pulled below this equilibrium position and released, what is its period of oscillation? (the spring stretches to 30.0 cm)

Please, show me.

Answer 1

Answer 2

mg=kx

k=0.3*9.8/0.3 = 9.8 Nm

time period = 1/2pi sqrt(K/m)

=1/2pisqrt(9.8/0.3) =0.91s

Answer 3

Spring force= gravitational force

kx=mg=> k=0.3*9.8/0.3=9.8 N/m

where x is the stretch=30cm=0.3m

m=0.3kg

w=sqrt(k/m)=(9.8/0.3)^(0.5)=5.71 rad

Time period T=2*pi/w=2*pi/5.71=1.1 s