Question

Can you please have this solve? I am Reviewing for a final thank you

Answer 1

Part A. We are given that the deformation of a string from its equilibrium position is given by the following equation:

$r=2\cos\theta+\sqrt{3}$

To determine the value of the angle for which the spring in at equilibrium we need to set the equation to zero:

$2\cos\theta+\sqrt{3}=0$

Now, we solve for the time. First, we subtract the square root of 3 to both sides:

$2\cos\theta=-\sqrt{3}$

Now, we divide both sides by 2:

$\cos\theta=\frac{-\sqrt{3}}{2}$

now, we take the inverse function of the cosine:

$\theta=cos^{-1}(\frac{-\sqrt{3}}{2})$

Solving the operation:

$\theta=\frac{5\pi}{6},\frac{7\pi}{6}$

This two angles apply for the interval of time:

$(0,2\pi)$

Part B. Now, we are asked to double the angle. We get the following equation:

$r=2\cos(2\theta)+\sqrt{3}$

Now, we set the equation to zero:

$2\cos(2\theta)+\sqrt{3}=0$

Now, we solve for the time similarly as we did in the previous point:

$2\theta=cos^{-1}(\frac{-\sqrt{3}}{2})$

Now, we solve the right side:

$2\theta=\frac{5\pi}{6},\frac{7\pi}{6}$

Dividing both sides by 2:

$\theta=\frac{5\pi}{12},\frac{7\pi}{12}$

In this case, for the interval between 0 and 2 pi we have that the possibe values of time are:

$\theta=\frac{5\pi}{12},\frac{7\pi}{12},\frac{19\pi}{12},\frac{17\pi}{12}$

The difference with the values in part A is that the spring now passes through the equolibrium point at a greater frequency.

Part C. We are given that another spring has the following equation :

$g(\theta)=1-sin^2\theta+\sqrt{3}$

To determine the value of time where the spring are at equilibrium we set both equations equal:

$2\cos\theta+\sqrt{3}=1-sin^2\theta+\sqrt{3}$

We can cancel out the square root of 3:

$2\cos(\theta)=1-s\imaginaryI n^2\theta$

Now, we use the following trigonometric identity:

$\cos^2\theta=1-\sin^2\theta$

Substituting we get:

$2\cos\theta=\cos^2\theta$

Now, we subtract the left side:

$\cos^2\theta-2\cos\theta=0$

Now, we take cosine as common factor:

$\cos\theta(\cos\theta-2)=0$

Now, we set each factor to zero:

$\begin{gathered} cos\theta=0 \\ \theta=n\pi-\frac{\pi}{2} \end{gathered}$

For:

$n=1,2,3,4...$

For the second factor we have:

$\begin{gathered} cos\theta-2=0 \\ \cos\theta=2 \\ No\text{ solutions} \end{gathered}$

Therefore, the times are the solution to the first factor.