A 75 g block is placed into oscillation as it hangs from a vertical spring. If the amplitude of the motion is 22 cm and the block is observed to have a maximum speed of 14 m/s (as it passes through the equilibrium position), what is the spring constant?
A) 300 N/m
B) 22 N/m
C) 4.8 N/m
D) 1200 N/m

Question

Quizplus · Accepted Answer

The spring constant can be found using the formula for the maximum speed in simple harmonic motion, $v_{max} = \sqrt{\frac{k}{m}}A$, where $v_{max}$ is the maximum speed, $k$ is the spring constant, $m$ is the mass, and $A$ is the amplitude. Rearranging for $k$, we get $k = \frac{m v_{max}^2}{A^2}$. Substituting $m = 0.075 \, kg$, $v_{max} = 14 \, m/s$, and $A = 0.22 \, m$, we find $k = 300 \, N/m$.

A 75 G Block Is Placed into Oscillation as It