Chapter 11, Problem 50P

To determine

To Find: The factor at which kinetic energy is going to change.

Answer to Problem 50P

The factor at which the bug's maximum Kinetic Energy is going to change is $0.5625$ .

Explanation of Solution

Given info:

Total distance, $d_1=6\text{ cm = }0.06\text{ m}$

Decreased distance, $d_2=4.5\text{ cm = }0.045\text{ m}$

Formula used:

Potential energy (=Kinetic energy) at the amplitude A is expressed by following formula.

  $E_T=\frac{1}{2}\times k\times A^2$

Here,

  $A$ is amplitude.

Calculation:

Comparing two maximum kinetic energy,

  $\begin{array}{l}\frac{K{E}_2}{K{E}_1}=\frac{\frac{1}{2}\times k\times A_2{}^2}{\frac{1}{2}\times k\times A_1{}^2}=\frac{A_2{}^2}{A_1{}^2}\\ \frac{K{E}_2}{K{E}_1}=\frac{\frac{\pi }{4}\times d_2{}^2}{\frac{\pi }{4}\times d_1{}^2}=\frac{d_2{}^2}{d_1{}^2}\\ \frac{K{E}_2}{K{E}_1}={\left(\frac{0.045}{0.06}\right)}^2=0.5625\end{array}$

As, it is clearly seen from the above equation, acceleration is directly proportional to distance, so if acceleration is equal to $0.5a_{\max }$ only when the distance is half the available distance that is $x=0.5x_0$ .

Conclusion:

Thus, the factor at which the bug's maximum Kinetic Energy is going to change is $0.5625$ .