A package is oscillating on a spring scale with a period of 4.60 s.At time t = 0.00 s the package has zero speed and is at x = 8.30 cm.At what time after t = 0.00 s will the package first be at x = 4.15 cm?
A)0.575 s
B)0.767 s
C)1.15 s
D)1.30 s
E)1.53 s

Question

Quizplus · Accepted Answer

The motion of the package can be described by simple harmonic motion (SHM), where the displacement x from the equilibrium position as a function of time t is given by $x(t) = A \cos(\omega t + \phi)$, where A is the amplitude, $\omega$ is the angular frequency, and $\phi$ is the phase constant. At $t = 0$, the package is at its maximum displacement (amplitude $A = 8.30$ cm), and since it has zero speed, this corresponds to the cosine function starting from its maximum value. Therefore, the phase constant $\phi = 0$, and the equation simplifies to $x(t) = A \cos(\omega t)$.Given that the period $T = 4.60$ s, we can find the angular frequency $\omega$ using the relation $\omega = \frac{2\pi}{T}$. The displacement $x = 4.15$ cm is half the amplitude, which corresponds to the cosine function being equal to 0.5 (since the full amplitude represents cosine of 0 or $\pi$, and half amplitude corresponds to $\cos^{-1}(0.5)$).To find the time when $x = 4.15$ cm, we solve $4.15 = 8.30 \cos(\omega t)$ or $\cos(\omega t) = 0.5$. This gives $\omega t = \frac{\pi}{3}$ or $\frac{5\pi}{3}$ for the first instance within the period, but since we start from maximum displacement and move towards the equilibrium (decreasing x), we use $\omega t = \frac{\pi}{3}$.Substituting $\omega = \frac{2\pi}{4.60}$ into $\omega t = \frac{\pi}{3}$ gives $t = \frac{T}{6}$. Calculating this gives $t = \frac{4.60}{6} = 0.767$ s, which is the first time the package will be at $x = 4.15$ cm after $t = 0.00$ s.

A Package Is Oscillating on a Spring Scale with a Period