1. The problem statement, all variables and given/known data A 775 g mass is hung on a spring. As a result the spring stretches 20.5 cm. If the object is then pulled an additional 3.0 cm downward and released, what is the period of the resulting oscillation? 2. Relevant equations T = 2pi sqr root(m/k) Hooke's Law Fs=kx 3. The attempt at a solution For the period I know you use T = 2pi sqr root(m/k) So you must find k using Hooke's Law, which we find force (m*g) and total distance is our number plus the additional 3cm, so to solve for k, its k=F/x, which you then substitute into the original equation, but I don't think my answer is right. I'm assuming my x is wrong? Do you not add the two values? .775kg 20.5cm + 3cm = .235m Fs=kx (.775)(9.81) / .235m = 32.34 T = 2pi sqrroot (.775kg/32.34) = .97 97 seconds.