Chapter 11, Problem 24P

(a)

To determine

To Find: The displacement.

Answer to Problem 24P

The equation of displacement as a function of time is $y=0.28\sin (34.25)t$ .

Explanation of Solution

Given:

Amplitude, $A=0.28\text{ m}$

Spring constant, $k=305\text{ N/m}$

Mass, $m=0.260\text{ kg}$

Formula used:

Angular speed is required to determine the equation of displacement as a function of time. Mass has zero displacement and positive velocity at t = 0, which has the equation of sine function.

  $\begin{array}{l}y=A\sin (\omega t)\\ \omega =\sqrt{\frac{k}{m}}\end{array}$

Calculation:

Substitute the given values in the above equation,

  $\begin{array}{l}\omega =\sqrt{\frac{305}{0.26}}=34.25\text{ rad/s}\\ y=A\sin (\omega t)=0.28\sin (34.25)t\end{array}$

Conclusion:

Thus, equation of displacement as a function of time is $y=0.28\sin (34.25)t$

(b)

To determine

To Find: The time.

Answer to Problem 24P

The maximum extension of spring is $t_{\max }=0.04575+(n\times 0.183)$ and minimum extension of the spring is $t_{\min }=0.137+(n\times 0.183)$ .

Explanation of Solution

Given:

Angular speed $\omega =34.25\text{ rad/s}$

Formula used:

Period (T) is required to find out to determine the extension in spring, maximum extension occurs at quarter of the time period.

  $\begin{array}{l}T=\frac{2\pi }{\omega }\\ t_{\max }=\frac{T}{4}+(n\times t)\end{array}$

Calculation:

Substitute the given values to get the maximum tension in spring.

  $\begin{array}{l}T=\frac{2\pi }{34.25}=0.183\text{ s}\\ t_{\max }=\frac{0.183}{4}+(n\times 0.183)\\ t_{\max }=0.04575+(n\times 0.183)\\ \text{n is any whole number such as 0,1,2,3 and so on}\text{.}\end{array}$

Minimum extension occurs at the equilibrium point and it is at third quarter of time period.

  $t_{\min }=\frac{3\times T}{4}+(n\times t)$

Substitute the given values in the above equation,

  $\begin{array}{l}{t}_{\min }=\frac{3\times 0.183}{4}+(n\times 0.183)\\ t_{\max }=0.137+(n\times 0.183)\\ \text{n is any whole number such as 0,1,2,3 and so on}\text{.}\end{array}$

Conclusion:

Thus, the maximum extension of spring is $t_{\max }=0.04575+(n\times 0.183)$ and minimum extension of the spring is $t_{\min }=0.137+(n\times 0.183)$ .