Question

As in Problem 14, let the displacements be ${\mathit{y}}_{\mathbf{1}}\mathbf{=}\mathbf{3}\mathit{s}\mathit{i}\mathit{n}\left(\frac{t}{\sqrt{2}}\right)$ and${\mathit{y}}_{\mathbf{2}}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{t}$ . The pendulums start together at t = 0. Make computer plots to estimate when they will be together again and then, by computer, solve the equation ${\mathbf{y}}_{\mathbf{1}}\mathbf{=}{\mathbf{y}}_{\mathbf{2}}$ for the root near your estimate.

Short answer

The time taken by both the pendulums with displacement functions are$3\sin \left(\frac{t}{\sqrt{2}}\right)$ and$\text{sin t}$ to come at the same position are 4.913 and 9.056 seconds.

Step-by-step solution

Definition of amplitude, period, frequency, and velocity amplitude.

The wave equation in standard form is $\text{S = Acos(}\omega \text{t +}\varphi )$ or$\text{S = Asin(}\omega \text{t +}\varphi )$

Where, A is amplitude;

$\omega =\frac{2\pi }{\text{T}}$Angular frequency$\omega$and period T.

Period is defined as the time taken by a function to repeat itself after a definite interval.

For example, in$\mathrm{P}=\mathrm{sinx}$ period of the function is$2\pi$ .

Given parameters

Given that the displacement of two simple pendulums are $3\sin \left(\frac{t}{\sqrt{2}}\right)$ and sin t.

They are not together at t = 0.

Find out the time period at which both pendulums again are together at same position and solve for$y_1=y_2$ .

Sketch the graph of displacement v/s time for both pendulums

To draw the curve starts the curve with x = 0 . Both the functions have the amplitude of 3 and 1 respectively.

The time period for each displacement function can be calculated by using formulae:

$\omega =\frac{2\pi }{T}$

For the first pendulum,

$$\begin{gathered} \frac{1}{\sqrt{2}}=\frac{2\pi }{T} \\ T=2\sqrt{2}\pi \end{gathered}$$

For second pendulum,

$$\begin{gathered} 1=\frac{2\pi }{T} \\ T=2\pi \end{gathered}$$

Use the graphing utilities to sketch the curve of both simple pendulums as follows:

Here, x is the displacement and t is the time.

Solve for equation  y1=y2

The time required for each cycle for first pendulum is $2\sqrt{2}\pi$ seconds.

Also, time required for each cycle for second pendulum is $2\pi$ seconds.

${\text{y}}_{\text{1}}{\text{=y}}_{\text{2}}$

Then

$3\sin \left(\frac{t}{\sqrt{2}}\right)=\sin t$

On solving the equation graphically, the values of time at which both the functions have same values or amplitude are 4.913 second and 9.056 second.