A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an additional 5.0 cm and released. Its position as a function of time is approximately
A) y = −0.10 sin 9.9t
B) y = 0.10 cos 9.9t
C) y = −0.10 cos (9.9t + .1)
D) y = 0.10 sin (9.9t + 5)
E) y = −0.05 cos 9.9t

Question

Quizplus · Accepted Answer

The correct answer is E) y = −0.05 cos 9.9t. This is because the displacement from the equilibrium position is 5.0 cm (or 0.05 m) and the motion is simple harmonic. The negative sign indicates a phase difference due to the initial displacement direction. The angular frequency $\omega = \sqrt{\frac{k}{m}}$, where $k$ is the spring constant and $m$ is the mass. Given the stretch of 10 cm by a 5.0 kg mass, we can find $k$ using $mg = kx$, then calculate $\omega$. However, without calculating, the form of the answer matches the expected form for a cosine function representing simple harmonic motion starting from an initial displacement, with the amplitude matching the initial displacement.

A Body of Mass 5