A block with mass m =6.7 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.27 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.5 m/s. The block oscillates on the spring without friction.
What is the oscillation frequency?

Question

A block with mass m =6.7 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.27 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.5 m/s. The block oscillates on the spring without friction. 
What is the oscillation frequency?

anonymous · Answer

i am using $$F=1/2\pi \sqrt{k/m}$$
and k=243.43 (which is correct

anonymous · Answer

Hey, the initial push downward with a velocity of v is a red herring IMHO. The velocity only determines the kinetic energy of the ball on spring, which in turn determines the maximum displacement from the equilibrium position (when the entire kinetic energy transforms into potential energy) . Since you are not asked this, it is safe to assume this was a decoy and the answer simply is:$$f = \omega/2\pi = \frac{ 1 }{ 2 \pi } \sqrt{\frac{ k }{ m }}$$
The value of k you have determined yourself already. (A spring with a mass is motionless and the stretch is x=0.27m. This means the force on the spring is equal to the weight of the block,$$k.x = m.g$$so$$k = \frac{ mg }{ x }$$and the oscillation frequency is $$f = \frac{ 1 }{ 2 \pi }\sqrt{\frac{ m.g }{ x.m }} = \frac{ 1 }{ 2 \pi }\sqrt{\frac{ g }{ x }} = \frac{ 1 }{ 2 \pi }\sqrt{\frac{ 9.81 m/s ^{2} }{ 0.27 m }} = 0.959 Hz$$