A 0.5-kg mass is suspended from a massless spring that has a force constant of 79 N/m.The mass is displaced 0.1m down from its equilibrium position and released.If the downward direction is negative,the displacement of the mass as a function of time is given by
A)y = 0.1 cos(158t - $\pi$)
B)y = 0.2 cos(158t - $\pi$)
C)y = 0.1 cos(12.6t - $\pi$)
D)y = 0.2 cos(12.6t + $\pi$)
E)y = 0.1 cos(2t + $\pi$)

Question

Quizplus · Accepted Answer

The displacement of a mass on a spring can be described by the equation: y = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency, t is time, and φ is the phase angle.
Given that the mass is displaced 0.1m down from its equilibrium position, we have A = 0.1m. The force constant k = 79 N/m, which can be used to find the angular frequency: 
ω = sqrt(k/m) = sqrt(79/0.5) = 12.6 rad/s
Thus, the displacement equation becomes: y = 0.1 cos(12.6t + φ).
We need to determine the phase angle φ. Since the mass is released from rest, we know that the displacement is at its maximum at t = 0, and cosine is at its maximum when the angle is 0. Therefore, we set y = 0.1 and solve for φ:
0.1 = 0.1 cos(φ)
cos(φ) = 1 
φ = 0
Therefore, the displacement equation is: y = 0.1 cos(12.6t), which is equivalent to choice C.

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