Chapter 11, Problem 19P

(a)

To determine

To Find:The general equation of pumpkin's position.

Answer to Problem 19P

The general equation which gives pumpkin's position $y$ as a function of time $t$ is $y=0.18\cos (9.7t)$ .

Explanation of Solution

Given:

Mass $m=2\text{ kg}$ .

Amplitude $A=18\text{ cm }=0.18\text{ m}$ .

Period $T=0.65\text{ s}$ .

Formula used:

Pumpkin's position is in positive $y$ direction and this position is just because of its compressive nature. Socosine function of the Pumpkin's position is derived by following way.

  $y=A\cos (\omega t)=A\cos \left(\frac{2\times \pi \times t}{T}\right)$

Here,

  $A$ is amplitude of spring,

  $T$ is time period of spring,

  $t$ is time.

Calculation:

Substitute the given values in the above equation,

  $y=0.18\cos \left(\frac{2\pi t}{0.65}\right)=0.18\cos (9.7\times t)$

Conclusion:

Thus, the general equation giving the pumpkin's position $y$ as a function of time $t$ is $y=0.18\cos (9.7t)$ .

(b)

To determine

To Find:The time required.

Answer to Problem 19P

The time required for the pumpkin to get back to its equilibrium position for the first time is $0.16\mathrm{s}$ .

Explanation of Solution

Given:

Period $T=0.65\text{ s}$ .

Formula used:

Time required by the pumpkin to get back to its equilibrium position is one quarter of the period, which is expressed by the following formula.

  $t=\frac{1}{4}\times T$

Here,

  $T$ is time period of spring.

Calculation:

Substitute the given values,

  $\text{time }t=\frac{1}{4}\times 0.65=0.16\text{ s}$

Conclusion:

Thus, the time required for the pumpkin to get back to its equilibrium position for the first time is $0.16\text{ s}$ .

(c)

To determine

To Find:The maximum speed.

Answer to Problem 19P

The maximum speed of the pumpkin is $1.74\text{ m/s}$ .

Explanation of Solution

Given:

Amplitude $A=18\text{ cm }=0.18\text{ m}$ .

Period $T=0.65\text{ s}$ .

Formula used:

The maximum speed of the pumpkin is determined by the following formula,

  $v_{\max }=\omega \times A=\left(\frac{2\pi }{T}\right)\times A$

Here,

  $\omega$ is angular frequency,

  $A$ is amplitude of spring,

  $T$ is time period of spring.

Calculation:

Substitute the given values,

  $\begin{array}{l}{v}_{\max }=\left(\frac{2\pi }{T}\right)\times A=\left(\frac{2\pi }{0.65}\right)\times 0.18\\ v_{\max }=1.74\text{ m/s}\end{array}$

Conclusion:

Thus, the maximum speed of the pumpkin is $1.74\text{ m/s}$ .

(d)

To determine

To Find:The maximum acceleration.

Answer to Problem 19P

The maximum acceleration of the pumpkin is $a_{\max }=17{\text{ m/s}}^{\text{2}}$ and it is attained first at the release point of pumpkin.

Explanation of Solution

Given:

Amplitude $A=18\text{ cm }=0.18\text{ m}$ .

Period $T=0.65\text{ s}$ .

Formula used:

The maximum acceleration of the pumpkin is determined by the following formula,

  $\begin{array}{l}{a}_{\max }={\omega }^2\times A\\ a_{\max }={\left(\frac{2\pi }{T}\right)}^2\times A\end{array}$

Here,

  $\omega$ is angular frequency,

  $A$ is amplitude of spring,

  $T$ is time period of spring.

Calculation:

Substitute the given values,

  $\begin{array}{l}{a}_{\max }={\left(\frac{2\pi }{T}\right)}^2\times A=\frac{4{\pi }^2}{{(0.65)}^2}\times 0.18\\ a_{\max }=17{\text{ m/s}}^{\text{2}}\end{array}$

Conclusion:

Thus, the maximum acceleration of the pumpkin is $a_{\max }=17{\text{ m/s}}^{\text{2}}$ and it is attained first at the release point of pumpkin.