1. The problem statement, all variables and given/known data A massless spring with no mass attached to it hangs from the ceiling. Its length is 0.2 meter. A mass M is now hung on the lower end of the spring. Imagine supporting the mass with your hand so that the spring remains relaxed (that is, not expanded or compressed), then suddenly remove your supporting hand. The mass and spring oscillate. The lowest position of the mass during the oscillations is 0.1 m below the place it was resting when you supported it. (a) What is the frequency of oscillation? (b) What is the velocity when the mass is 0.05 m below its original resting place? 2. Relevant equations t = sqrt(2*d/g) g = 9.8 frequency = w/2pi w^2=g/l x=Acos(wt+Φ) 3. The attempt at a solution These problems always throw me off. I do not want to be thrown off this time around! Please check my work. I am somewhat confident in it. 1. To find frequency of oscillating we find w. w is easy enough by the equation w^2=g/l. Since the length is 0.2 and g = 9.8 this comes out to 7. question: I used the correct length correct? Can I use this equation? This was giving to me in my pendulum notes but we are working with a spring... Is this incorrect? 2. With w, it is easy to see frequency = w/2pi, so 7/2pi is thus the frequency and part a. 3. To find velocity we need the position function which is x = 0.1sin(7*t) or equally 0.1cos(7t+pi/2) 4. Take the derivative which becomes velocity = 0.7cos(7t) 5. Find t for 0.05m. t=sqrt(2*0.05/9.8) = 0.101 6. Plug t into velocity equation to get 0.7cos(7*0.101) = 0.6999 = 0.7m/s Did I use wrong equations and get the wrong answer? I am somewhat skeptical about my final velocity as it is saying that cos(0)=1 and so it takes on 0.7.