A 3-kg block, attached to a spring, executes simple harmonic motion according to x = 2cos(50t)where x is in meters and t is in seconds.The spring constant of the spring is:
A)1 N/m
B)100 N/m
C)150 N/m
D)7500 N/m
E)none of these

Question

Quizplus · Accepted Answer

The equation of motion for a simple harmonic oscillator is given by $x = A\cos(\omega t)$, where $A$ is the amplitude of motion, $\omega$ is the angular frequency, and $t$ is time. In this case, $A = 2$ m and $\omega = 50$ rad/s. The angular frequency $\omega$ is related to the mass $m$ and the spring constant $k$ by the formula $\omega = \sqrt{\frac{k}{m}}$. Rearranging this formula to solve for $k$ gives $k = m\omega^2$. Substituting $m = 3$ kg and $\omega = 50$ rad/s into this formula yields $k = 3 \times (50)^2 = 3 \times 2500 = 7500$ N/m.

A 3-Kg Block, Attached to a Spring, Executes Simple Harmonic