Chapter 6.4, Problem 59E

(a)

To determine

To Find: The equation that models the horizontal displacement of the bob if it is at the maximum distance at right when $t=0.$

Answer to Problem 59E

  $y=1.5\cos \left(t\sqrt{\frac{9.8}{6}}\right)$

Explanation of Solution

Given: $y=A\cos \left(t\sqrt{\frac{g}{l}}\right)$

  

Given: $y=A\cos \left(t\sqrt{\frac{g}{l}}\right)$ and length = $6m$ and maximum horizontal distance to the right is $1.5m$ .

  $\begin{array}{l}y=A\cos \left(t\sqrt{\frac{g}{l}}\right)\\ y=1.5\cos \left(t\sqrt{\frac{9.8}{6}}\right)\end{array}$

(b)

To determine

To Find: The location of the bob at $4\sec$ .

Answer to Problem 59E

  $0.6m$ to the right.

Explanation of Solution

Given: $y=A\cos \left(t\sqrt{\frac{g}{l}}\right)$

  

Given: $y=A\cos \left(t\sqrt{\frac{g}{l}}\right)$

  $\begin{array}{l}y=A\cos \left(t\sqrt{\frac{g}{l}}\right)\\ y=1.5\cos \left(4\sqrt{\frac{9.8}{6}}\right)\\ y\approx 0.6m\end{array}$

Thus the location of the bob is $0.6m$ to the right.

(c)

To determine

To Find: The location of the bob at $7.9\sec$ .

Answer to Problem 59E

  $1.2m$ to the left.

Explanation of Solution

Given: $y=A\cos \left(t\sqrt{\frac{g}{l}}\right)$

  

Given: $y=A\cos \left(t\sqrt{\frac{g}{l}}\right)$

  $\begin{array}{l}y=A\cos \left(t\sqrt{\frac{g}{l}}\right)\\ y=1.5\cos \left(7.9\sqrt{\frac{9.8}{6}}\right)\\ y\approx -1.2m\end{array}$

Thus the location of the bob is $1.2m$ to the left.