Chapter 2, Problem 40PQ

(a)

To determine

Draw a sketch for the problem.

Answer to Problem 40PQ

Sketch for the problem,

Explanation of Solution

In the problem, the particle attached to the vertical spring is pulled down and released.

Sketch for the problem,

(b)

To determine

Draw the position-time graph.

Answer to Problem 40PQ

The position-time graph,

Explanation of Solution

Write the expression for position of the particle.

  $\overrightarrow{y}=\left(y_0\cos \frac{2\pi t}{T}\right)\widehat{j}$

Here, $y_0$ is the amplitude of the particle.

The position-time graph,

(c)

To determine

Draw the velocity-time graph.

Answer to Problem 40PQ

The velocity-time graph,

Explanation of Solution

Write the expression for position of the particle.

  $\overrightarrow{y}=\left(y_0\cos \frac{2\pi t}{T}\right)\widehat{j}$

Here, $y_0$ is the amplitude of the particle.

Derivate the position of the particle to find the velocity of the particle

  $\overrightarrow{v}=\frac{d\overrightarrow{y}}{dt}$

Conclusion:

Substitute $\left(y_0\cos \frac{2\pi t}{T}\right)\widehat{j}$ for $\overrightarrow{y}$ to find $\overrightarrow{v}$.

  $\begin{array}{c}\overrightarrow{v}=\frac{d\left(\left(y_0\cos \frac{2\pi t}{T}\right)\widehat{j}\right)}{dt}\\ =\left(-y_0\frac{2\pi }{T}\sin \frac{2\pi t}{T}\right)\widehat{j}\end{array}$

The velocity-time graph,

(d)

To determine

Draw the acceleration-time graph.

Answer to Problem 40PQ

The acceleration -time graph,

Explanation of Solution

Write the expression for velocity of the particle.

  $\overrightarrow{v}=\left(-y_0\frac{2\pi }{T}\sin \frac{2\pi t}{T}\right)\widehat{j}$

Here, $y_0$ is the amplitude of the particle.

Derivate the velocity of the particle to find the acceleration of the particle

  $\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}$

Conclusion:

Substitute $\left(-y_0\frac{2\pi }{T}\sin \frac{2\pi t}{T}\right)\widehat{j}$ for $\overrightarrow{v}$ to find $\overrightarrow{a}$.

  $\begin{array}{c}\overrightarrow{a}=\frac{d\left(\left(-y_0\frac{2\pi }{T}\sin \frac{2\pi t}{T}\right)\widehat{j}\right)}{dt}\\ =\left(-y_0\frac{4{\pi }^2}{T^2}\cos \frac{2\pi t}{T}\right)\widehat{j}\end{array}$

The acceleration-time graph,

  

(e)

To determine

The time at which the speed of the particle is maximum and the position of the particle at maximum speed.

Answer to Problem 40PQ

The time at which the speed of the particle is maximum is $t=\frac{\pi }{4}T$ and $t=\frac{3\pi }{4}T$.

The particle will be at equilibrium position when it is at maximum speed.

Explanation of Solution

Write the expression for velocity of the particle.

  $\overrightarrow{v}=\left(-y_0\frac{2\pi }{T}\sin \frac{2\pi t}{T}\right)\widehat{j}$

Here, $y_0$ is the amplitude of the particle.

Conclusion:

The speed of the particle is maximum if $\sin \left(\frac{2\pi t}{T}\right)=1$. Thus, the time will be $t=\frac{\pi }{4}T$ and $t=\frac{3\pi }{4}T$. The position of the particle will be at $y=0$ when the speed of the particle is maximum.

(f)

To determine

The time at which the magnitude of acceleration of the particle is maximum and the position of the particle at maximum acceleration.

Answer to Problem 40PQ

The time at which the acceleration of the particle is maximum is $t=0$, $t=\frac{1}{2}T$ and $t=T$.

The particle will be at furthest away from equilibrium position when it is at maximum speed.

Explanation of Solution

Write the expression for velocity of the particle.

  $\overrightarrow{a}=\left(-y_0\frac{4{\pi }^2}{T^2}\cos \frac{2\pi t}{T}\right)\widehat{j}$

Here, $y_0$ is the amplitude of the particle.

Conclusion:

The acceleration of the particle is maximum if $\cos \left(\frac{2\pi t}{T}\right)=1$. Thus, the time will be $t=0$, $t=\frac{1}{2}T$ and $t=T$. The position of the particle will be at $y=\pm y_0$ when the acceleration of the particle is maximum.