Solve the problem.
-A pendulum of length $\mathrm { L }$, when displaced horizontally and released, oscillates with harmonic motion according to the equation $\mathrm { y } = \mathrm { A } \sin ( ( \sqrt { \mathrm { g } / \mathrm { L } } ) \mathrm { t } + \pi / 2 )$, where $\mathrm { y }$ is the distance in meters from the rest position $t$ seconds after release, and $g = 9.8 \mathrm {~m} / \mathrm { sec } ^ { 2 }$. Identify the period, amplitude, and phase shift when $\mathrm { A } = 0.39 \mathrm {~m}$ and $\mathrm { L } = 0.49 \mathrm {~m}$. Round all answers to the nearest hundredth.
A) $0.70 \mathrm { sec } , 0.39 \mathrm {~m} , - 0.70$ units to the left
B) $1.41 \mathrm { sec } , 0.39 \mathrm {~m} , 0.35$ units to the right
C) $0.31 \mathrm { sec } , 0.39 \mathrm {~m} , - 0.079$ units to the left
D) $1.41 \mathrm { sec } , 0.39 \mathrm {~m} , - 0.35$ units to the left

Question

Quizplus · Accepted Answer

The period of the pendulum is given by $T = 2\pi\sqrt{L/g}$, which calculates to approximately 1.41 seconds for $L = 0.49$ m. The amplitude $A = 0.39$ m is given directly. The phase shift, derived from the phase angle $\pi/2$, translates to a shift in time, which is calculated as $\Delta t = \frac{\phi}{\omega} = \frac{\pi/2}{\sqrt{g/L}}$, resulting in a shift to the left, approximately -0.35 units.

Solve the Problem L\mathrm { L }L , When Displaced Horizontally and Released, Oscillates with Harmonic Motion

Solve the Problem $\mathrm { L }$ , When Displaced Horizontally and Released, Oscillates with Harmonic Motion