Question

A horizontal spring-mass system oscillates on a frictionless table. If the ratio of the mass to the spring constant is 0.042 kg·m/N, and the maximum speed of the mass was measured to be 5.9 m/s, the maximum extension of the spring will be

Answer 1

$E = E_k+E_p=\text{const}$

$E_k=\frac{mV^2}{2}\ E_p=\frac{kx^2}{2}$

$\text{where } k \text{ is the stiffness of the spring x the size of compression (extension)}$

$\text{for }x_{max}\ V=0 \ E=\frac{kx^2_{max}}{2}$

$\text{for }x=0\ V_{max}\ E=\frac{mV_{max}^2}{2}$

$\frac{kx^2_{max}}{2}=\frac{mV^2_{max}}{2}$

$x_{max}=\sqrt{\frac{m}{k}}*V=\sqrt{0.042}*5.9\approx1.21m$

Answer: $x_{max}\approx1.21m$