Question

A spring in simple harmonic motion is found to have a velocity of 5.65 m/s when a mass of 50.0 grams is placed on the end of the spring. If the spring constant is 12.5 N/m, what is the maximum

displacement from equilibrium position that the spring system experiences?

Answer 1

$\frac{k\times x_{max}^2}{2}=\frac{m\times v_{max}^2}{2};\\x_{max}=\sqrt{\frac{m\times v_{max}^2}{k}};\\x_{max}=\sqrt{\frac{0.05\times 5.65^2}{12/5}} =0.357m$

Answer: 0.357m