Chapter 14, Problem 34P

(a)

To determine

The distance travelled by the particle during the time $t=0\text{s}\text{to}2\text{s}$ .

Explanation of Solution

Given:

The time period of the particle is $8.0\text{s}$ .

The amplitude of the oscillation of the particle is $12\text{cm}$ .

Formula used:

The position of the particle is given as:

  $x=A\cos \left(\omega t+\delta \right)$   ............. (1)

Here, $x$ is the position of the particle, $A$ is the amplitude, $\omega$ is the angular frequency, $t$ is the time and $\delta$ is the phase constant.

Write the expression for the angular frequency of oscillation.

  $\omega =\frac{2\pi }{T}$   ............. (2)

Substitute $8.0\text{s}$ for $T$ in equation (2)

  $\begin{array}{l}\omega =\frac{2\pi }{8.0\text{s}}\\ \omega =\frac{\pi }{4}{\text{s}}^{\text{-1}}\end{array}$

Write the expression for the initial position of the particle with amplitude and phase constant.

  $x_0=A\cos \delta$

Simplify the above equation we get.

  $\delta ={\cos }^{-1}\left(\frac{x_0}{A}\right)$   ............. (3)

Substitute $0$ for $x_0$ in equation (3).

  $\begin{array}{l}\delta ={\cos }^{-1}\left(\frac{0}{A}\right)\\ \delta =\frac{\pi }{2}\end{array}$

Substitute $\frac{\pi }{2}$ for $\delta$ , $12\text{cm}$ for $A$ and $\frac{\pi }{4}{\text{s}}^{\text{-1}}$ for $\omega$ in equation (1)

  $\begin{array}{l}x=A\cos \left(\omega t+\delta \right)\\ x=\left(12\text{cm}\right)\cos \left(\frac{\pi }{4}{\text{s}}^{\text{-1}}t+\frac{\pi }{2}\right)\end{array}$

Now the distance particle travels in initial time $t_{\text{initial}}$ and final time $t_{\text{final}}$ is:

  $x=\left(12\text{cm}\right)\cos \left[\left(\frac{\pi }{4}{s}^{-1}\right)t_{\text{final}}+\frac{\pi }{2}\right]-\left(\left(12\text{cm}\right)\cos \left[\left(\frac{\pi }{4}{s}^{-1}\right)t_{\text{initial}}+\frac{\pi }{2}\right]\right)$   ............. (4)

Calculation:

Substitute $2.0\text{s}$ for $t_{\text{final}}$ and $0\text{s}$ for $t_{\text{initial}}$ in equation (4).

  $\begin{array}{l}x=\left(12\text{cm}\right)\cos \left[\left(\frac{\pi }{4}{s}^{-1}\right)2.0\text{s}+\frac{\pi }{2}\right]-\left(\left(12\text{cm}\right)\cos \left[\left(\frac{\pi }{4}{s}^{-1}\right)0\text{s}+\frac{\pi }{2}\right]\right)\\ x=\left(12\text{cm}\right)\cos \left[\left(\frac{\pi }{4}{s}^{-1}\right)2.0\text{s}+\frac{\pi }{2}\right]-\left(\left(12\text{cm}\right)\cos \frac{\pi }{2}\right)\\ x=12\text{cm}\end{array}$

Conclusion:

The distance the particle travels at $t=0\text{s}\text{to}2\text{s}$ is $x=12\text{cm}$ .

(b)

To determine

The distance travelled by the particle at time $t=0.2\text{s}\text{to}0.4\text{s}$.

Explanation of Solution

Given:

The time period of the particle is $8.0\text{s}$ .

The amplitude of the oscillation of the particle is $12\text{cm}$ .

Formula used:

The position of the particle is given as:

  $x=A\cos \left(\omega t+\delta \right)$   ............. (1)

Here, $x$ is the position of the particle, $A$ is the amplitude, $\omega$ is the angular frequency, $t$ is the time and $\tau$ is the phase constant.

Calculation:

Substitute $4.0\text{s}$ for $t_{\text{final}}$ and $2.0s$ for $t_{\text{initial}}$ in equation (4).

  $\begin{array}{l}x=\left(12\text{cm}\right)\cos \left[\left(\frac{\pi }{4}{s}^{-1}\right)4.0\text{s}+\frac{\pi }{2}\right]-\left(\left(12\text{cm}\right)\cos \left[\left(\frac{\pi }{4}{s}^{-1}\right)\text{2s}+\frac{\pi }{2}\right]\right)\\ x=\left(12\text{cm}\right)\cos \left[\frac{3\pi }{2}\right]-\left(12\text{cm}\right)\cos (\pi )\\ x=12\text{cm}\end{array}$

Conclusion:

The position of the particle is $12\text{cm}$ .

(c)

To determine

The distance travelled by the particle $t=0\text{s}\text{to}1\text{s}$ .

Explanation of Solution

Given:

The time period of the particle is $0.8\text{s}$ .

The amplitude of the oscillation of the particle is $12\text{cm}$ .

Formula used:

The position of the particle is given as:

  $x=A\cos \left(\omega t+\delta \right)$

Calculation:

Substitute $1\text{s}$ for $t_{\text{final}}$ and $0\text{s}$ for $t_{\text{initial}}$ in equation (4).

  $\begin{array}{l}x=\left(12\text{cm}\right)\cos \left[\left(\frac{\pi }{4}{s}^{-1}\right)1.0\text{s}+\frac{\pi }{2}\right]-\left(\left(12\text{cm}\right)\cos \left[\left(\frac{\pi }{4}{s}^{-1}\right)\text{0s}+\frac{\pi }{2}\right]\right)\\ x=\left(12\text{cm}\right)\cos \left(\frac{3\pi }{4}\right)\\ x=-8.48\text{cm}\end{array}$

Conclusion:

The position of the particle is $-8.48\text{cm}$ .

(d)

To determine

The distance travelled by the particle $t=\text{1}\text{.0s}\text{to}\text{2}\text{.0s}$ .

Explanation of Solution

Given:

The time period of the particle is $0.8\text{s}$ .

The amplitude of the oscillation of the particle is $12\text{cm}$ .

Formula used:

The position of the particle is given as:

  $x=A\cos \left(\omega t+\delta \right)$

Calculation:

Substitute $2.0\text{s}$ for $t_{\text{final}}$ and $1.0\text{s}$ for $t_{\text{initial}}$ in equation (4).

  $\begin{array}{l}x=\left(12\text{cm}\right)\cos \left[\left(\frac{\pi }{4}{s}^{-1}\right)\text{2}\text{.0s}+\frac{\pi }{2}\right]-\left(\left(12\text{cm}\right)\cos \left[\left(\frac{\pi }{4}{s}^{-1}\right)\text{1}\text{.0s}+\frac{\pi }{2}\right]\right)\\ x=\left(12\text{cm}\right)\cos \left[\pi \right]-\left(\left(12\text{cm}\right)\cos \left[\frac{3\pi }{4}\right]\right)\\ x=-3.52\text{cm}\end{array}$

Conclusion:

The position of the particle is $-3.52\text{cm}$ .