Question

Suppose a pendulum of length L meters makes an angle of θ radians with the vertical, as in the figure. It can be shown that as a function of time, satisfies the differential equation dt2d2θ​+Lg​sinθ=0 where g=9.8 m/s2 is the acceleration due to gravity. For θ near zero we can use the linear approximation sin(θ)≈θ to get a linear differential equation dt2d2θ​+Lg​θ=0 Use the linear differential equation to answer the following questions. (a) Determine the equation of motion for a pendulum of length 2 meters having initial angle 0.2 radians and initial angular velocity dtdθ​=0.4 radians per second.

Answer 1

The equation of motion for a pendulum of length 2 meters with an initial angle of 0.2 radians and initial angular velocity of 0.4 radians per second is dt2d2θ​ + 1.96θ = 0.

Step-by-step explanation:

To find the equation of motion for a pendulum of length 2 meters with an initial angle of 0.2 radians and initial angular velocity of 0.4 radians per second, we can use the linear differential equation dt2d2θ​ + Lg​θ = 0, where L is the length of the pendulum and g is the acceleration due to gravity.

Plugging in the values L = 2 and θ = 0.2 gives us dt2d2θ​ + (9.8 m/s^2)(0.2) = 0.

This simplifies to dt2d2θ​ + 1.96θ = 0, which is the equation of motion for the given pendulum.

Answer 2

1. The equation of motion for the pendulum is:

`\( \theta(t) = 0.4\cos(\omega t) + \frac{0.3}{\sqrt{(g)/(L)}}\sin(\omega t) \).`

2. The period of the pendulum is
`\( 1.42 \)` seconds.

Solve the linear differential equation
`\( \frac{{d^2\theta}}{{dt^2}} + (g)/(L)\theta = 0 \)` to find the equation of motion for the given initial conditions.

The general solution to this differential equation is
`\( \theta(t) = A\cos(\omega t) + B\sin(\omega t) \), where \( \omega = \sqrt{(g)/(L)} \).`

Given the initial conditions:

`\( \theta(0) = 0.4 \) radians and \( \frac{{d\theta}}{{dt}}(0) = 0.3 \) radians/second.`

From the initial condition
`\( \theta(0) = 0.4 \),` we get:

`\( \theta(0) = A\cos(0) + B\sin(0) = A = 0.4 \).`

From the initial condition
`\( \frac{{d\theta}}{{dt}}(0) = 0.3 \),` we derive the first derivative of
`\( \theta(t) \):`

`\( \frac{{d\theta}}{{dt}} = -A\omega\sin(\omega t) + B\omega\cos(\omega t) \),`

`\( \frac{{d\theta}}{{dt}}(0) = -0.4\omega\sin(0) + B\omega\cos(0) = B\omega = 0.3 \).`

So,
`\( B = (0.3)/(\omega) = \frac{0.3}{\sqrt{(g)/(L)}} \).`

Therefore, the equation of motion for the pendulum is:

`\( \theta(t) = 0.4\cos(\omega t) + \frac{0.3}{\sqrt{(g)/(L)}}\sin(\omega t) \).`

To find the period of the pendulum, the period
`\( T \)` is given by
`\( T = (2\pi)/(\omega) \).`

So,
`\( T = \frac{2\pi}{\sqrt{(g)/(L)}} \).`

Given that
`\( g = 9.8 \, \text{m/s}^2 \)` and
`\( L = 0.5 \) meters`, we can compute the period:

`\( T = \frac{2\pi}{\sqrt{(9.8)/(0.5)}} \approx (2\pi)/(√(19.6)) \approx (2\pi)/(4.43) \approx 1.42 \) seconds (approximately).`

Therefore, the period of the pendulum is approximately
`\( 1.42 \)` seconds.

The complete question is here:

Suppose a pendulum of length L meters makes an angle of θ radians with the vertical, as in the figure. It can be shown that as a function of time, θ satisfies the differential equation

where
`g=9.8 m/s2` is the acceleration due to gravity. For θ near zero we can use the linear approximation sin(θ)≈θ to get a linear differential equation

`$(d^2 \theta)/(d t^2)+(g)/(L) \theta=0$`

Use the linear differential equation to answer the following questions.

(a) Determine the equation of motion for a pendulum of length 0.5 meters having initial angle 0.4 radians and initial angular velocity
`$(d \theta)/(d t)=0.2$` radians per second.

θ(t)= radians

(b) What is the period of the pendulum? That is, what is the time for one swing back and forth?