The figure shows a pendulum with length L and the angle $\theta$ from the vertical to the pendulum. It can be shown that $\theta$ , as a function of time, satisfies the nonlinear differential equation $\frac { d ^ { 2 } \theta } { d t ^ { 2 } } + \frac { g } { L } \sin \theta = 0$ where $g = 9.8 \mathrm {~m} / \mathrm { s } ^ { 2 }$ $\text { is the acceleration due to gravity. For small values of }$ $\theta$ we can use the linear approximation $\sin \theta = \theta$ $\text { and then the differential equation becomes linear. Find the equation }$ $\text { of motion of a pendulum with length } 1 \mathrm {~m} \text { if } \theta \text { is initially } 0.2 \mathrm { rad } \text { and the initial }$ $\text { angular velocity is }$ $\frac { d \theta } { d t } = 1 \mathrm { rad } / \mathrm { s }$
A) $$\theta ( t ) = 0.2 \cos ( \sqrt { 9.8 } t ) + \frac { 1 } { \sqrt { 9.8 } } \sin ( \sqrt { 9.8 } t )$$
B) $\theta ( t ) = 0.2 \cos ( \sqrt { 9.8 } t ) + 2 \sin ( \sqrt { 9.8 } t )$
C) $\theta ( t ) = 2 \cos ( 9.8 t ) + \frac { 1 } { 9.8 } \sin ( 9.8 t )$
D) $\theta ( t ) = \frac { 1 } { 9.8 } \cos ( \sqrt { 9.8 } t ) + 0.2 \sin ( \sqrt { 9.8 } t )$
E) $\theta ( t ) = 0.2 \sin ( \sqrt { 9.8 } t ) + \frac { 1 } { \sqrt { 9.8 } } \cos ( \sqrt { 9.8 } t )$

Question

The figure shows a pendulum with length L and the angle $\theta$ from the vertical to the pendulum. It can be shown that $\theta$ , as a function of time, satisfies the nonlinear differential equation $\frac { d ^ { 2 } \theta } { d t ^ { 2 } } + \frac { g } { L } \sin \theta = 0$ where $g = 9.8 \mathrm {~m} / \mathrm { s } ^ { 2 }$ $\text { is the acceleration due to gravity. For small values of }$ $\theta$ we can use the linear approximation $\sin \theta = \theta$ $\text { and then the differential equation becomes linear. Find the equation }$ $\text { of motion of a pendulum with length } 1 \mathrm {~m} \text { if } \theta \text { is initially } 0.2 \mathrm { rad } \text { and the initial }$ $\text { angular velocity is }$ $\frac { d \theta } { d t } = 1 \mathrm { rad } / \mathrm { s }$ 
A) $$\theta ( t ) = 0.2 \cos ( \sqrt { 9.8 } t ) + \frac { 1 } { \sqrt { 9.8 } } \sin ( \sqrt { 9.8 } t )$$ 
B) $\theta ( t ) = 0.2 \cos ( \sqrt { 9.8 } t ) + 2 \sin ( \sqrt { 9.8 } t )$ 
C) $\theta ( t ) = 2 \cos ( 9.8 t ) + \frac { 1 } { 9.8 } \sin ( 9.8 t )$ 
D) $\theta ( t ) = \frac { 1 } { 9.8 } \cos ( \sqrt { 9.8 } t ) + 0.2 \sin ( \sqrt { 9.8 } t )$ 
E) $\theta ( t ) = 0.2 \sin ( \sqrt { 9.8 } t ) + \frac { 1 } { \sqrt { 9.8 } } \cos ( \sqrt { 9.8 } t )$

Quizplus · Accepted Answer

The Answer of The figure shows a pendulum with length L and the angle $\theta$ from the vertical to the pendulum. It can be shown that $\theta$ , as a function of time, satisfies the nonlinear differential equation $\frac { d ^ { 2 } \theta } { d t ^ { 2 } } + \frac { g } { L } \sin \theta = 0$ where $g = 9.8 \mathrm {~m} / \mathrm { s } ^ { 2 }$ $\text { is the acceleration due to gravity. For small values of }$ $\theta$ we can use the linear approximation $\sin \theta = \theta$ $\text { and then the differential equation becomes linear. Find the equation }$ $\text { of motion of a pendulum with length } 1 \mathrm {~m} \text { if } \theta \text { is initially } 0.2 \mathrm { rad } \text { and the initial }$ $\text { angular velocity is }$ $\frac { d \theta } { d t } = 1 \mathrm { rad } / \mathrm { s }$ 
A) $$\theta ( t ) = 0.2 \cos ( \sqrt { 9.8 } t ) + \frac { 1 } { \sqrt { 9.8 } } \sin ( \sqrt { 9.8 } t )$$ 
B) $\theta ( t ) = 0.2 \cos ( \sqrt { 9.8 } t ) + 2 \sin ( \sqrt { 9.8 } t )$ 
C) $\theta ( t ) = 2 \cos ( 9.8 t ) + \frac { 1 } { 9.8 } \sin ( 9.8 t )$ 
D) $\theta ( t ) = \frac { 1 } { 9.8 } \cos ( \sqrt { 9.8 } t ) + 0.2 \sin ( \sqrt { 9.8 } t )$ 
E) $\theta ( t ) = 0.2 \sin ( \sqrt { 9.8 } t ) + \frac { 1 } { \sqrt { 9.8 } } \cos ( \sqrt { 9.8 } t )$

The Figure Shows a Pendulum with Length L and the Angle