Chapter 2, Problem 114P

(a)

To determine

To plot:The position x of a body oscillating on a spring as a function of time.

Explanation of Solution

Given:

The equation of position x as a function of time

  $x=A\sin \omega t$

The values of the constants

  $\begin{array}{l}A=5.0\text{ cm}\\ \omega =0.175{\text{ s}}^{-1}\end{array}$

The time interval

  $0\le t\le 36\text{ s}$

Calculation:

Using the given values of the variables in the given equation,

  $x=\left(5.0\text{ cm}\right)\sin \left(0.175{\text{ s}}^{-1}\right)t$  .....(1)

On a spreadsheet calculate the values of the position with respect to time and plot a graph as shown.

t in s	x in cm
0	0
1	0.87054
2	1.71449
3	2.50607
4	3.22109
5	3.83772
6	4.33712
7	4.70403
8	4.92725
9	4.99996
10	4.91993
11	4.68961
12	4.31605
13	3.81064
14	3.18882
15	2.4696
16	1.67494
17	0.82912
18	-0.042
19	-0.9119
20	-1.7539
21	-2.5424
22	-3.2531
23	-3.8645
24	-4.3579
25	-4.7181
26	-4.9342
27	-4.9996
28	-4.9123
29	-4.6749
30	-4.2947
31	-3.7833
32	-3.1563
33	-2.433
34	-1.6353
35	-0.7876
36	0.08407

  

Figure 1

Conclusion:

Thus, the position x of the object which undergoes oscillation following the equation $x=\left(5.0\text{ cm}\right)\sin \left(0.175{\text{ s}}^{-1}\right)t$ is graphed as a function of time.

(b)

To determine

To measure:The slope of the $x-t$ graph at $t=0$ and determine the value of the velocity at the instant of time.

Answer to Problem 114P

The velocity of the object at time $t=0$ is equal to the slope of the graph at $t=0$ and is found to be $0.9\text{ cm/s}$ .

Explanation of Solution

Given:

The $x-t$ graph shown in Figure 1 .

Calculation:

Draw a tangent to the curve at time $t=0$ and measure the slope of the tangent.

  

Figure 2

From Figure 2, the slope of the tangent (drawn in red) is given by,

  $\begin{array}{c}\text{slope}=\frac{1.8\text{ cm}-0}{2.0\text{ s}-0}\\ =0.90\text{ cm/s}\end{array}$

Hence the velocity of the object at time $t=0$ is equal to,

  $v_x=0.90\text{ cm/s}$

Conclusion:

Thus, the velocity of the object at time $t=0$ is equal to the slope of the graph at $t=0$ and is found to be $0.9\text{ cm/s}$ .

(c)

To determine

To calculate:The average velocity for a series of intervals starting from $t=0$ and ending at $t=6.0,3.0,2.0,1.0,0.50\&0.25\text{ s}$ .

Answer to Problem 114P

The average velocities for the time intervals starting at $t=0$ and ending at $t=6.0,3.0,2.0,1.0,0.50\&0.25\text{ s}$ are $0.72\text{ cm/s, 0}\text{.86 cm/s, 0}\text{.86 cm/s, 0}\text{.87 cm/s, 0}\text{.87 cm/s and 0}\text{.87 cm/s}$ respectively.

Explanation of Solution

Given:

The equation for the position of the oscillating particle

  $x=\left(5.0\text{ cm}\right)\sin \left(0.175{\text{ s}}^{-1}\right)t$

The times at which the average velocity is determined

  $t=6.0,3.0,2.0,1.0,0.50\&0.25\text{ s}$

Formula used:

The average velocity of a particle is the rate of change of position of the object during the time interval.

  $\overline{v}=\frac{x-x_0}{\Delta t}$  .....(2)

Calculation:

Determine the value of the position of the object $x_0$ at time $t=0$ by substituting the value of t in equation (1).

  $\begin{array}{c}{x}_0=\left(5.0\text{ cm}\right)\left[\sin \left(0.175{\text{ s}}^{-1}\right)t\right]\\ =\left(5.0\text{ cm}\right)\left[\sin \left(0.175{\text{ s}}^{-1}\right)\left(0\text{ s}\right)\right]\\ =0\end{array}$

Determine the position of the particle at time $t=6\text{ s}$ .

  $\begin{array}{c}{x}_6=\left(5.0\text{ cm}\right)\left[\sin \left(0.175{\text{ s}}^{-1}\right)t\right]\\ =\left(5.0\text{ cm}\right)\left[\sin \left(0.175{\text{ s}}^{-1}\right)\left(6.0\text{ s}\right)\right]\\ =4.34\text{ cm}\end{array}$

Determine the average velocity for the time interval $t=0$ to $t=6\text{ s}$

  $\begin{array}{c}{\overline{v}}_6=\frac{x_6-x_0}{\Delta t}\\ =\frac{4.34\text{ cm}-0}{6.0\text{ s}-0}\\ =0.72\text{ cm/s}\end{array}$

Find the position of the particle at time $t=3.0\text{ s}$ .

  $\begin{array}{c}{x}_3=\left(5.0\text{ cm}\right)\left[\sin \left(0.175{\text{ s}}^{-1}\right)t\right]\\ =\left(5.0\text{ cm}\right)\left[\sin \left(0.175{\text{ s}}^{-1}\right)\left(3.0\text{ s}\right)\right]\\ =2.51\text{ cm}\end{array}$

Determine the average velocity for the time interval $t=0$ to $t=3.0\text{ s}$

  $\begin{array}{c}{\overline{v}}_3=\frac{x_3-x_0}{\Delta t}\\ =\frac{2.51\text{ cm}-0}{3.0\text{ s}-0}\\ =0.86\text{ cm/s}\end{array}$

Find the position of the particle at time $t=2.0\text{ s}$ .

  $\begin{array}{c}{x}_2=\left(5.0\text{ cm}\right)\left[\sin \left(0.175{\text{ s}}^{-1}\right)t\right]\\ =\left(5.0\text{ cm}\right)\left[\sin \left(0.175{\text{ s}}^{-1}\right)\left(2.0\text{ s}\right)\right]\\ =1.71\text{ cm}\end{array}$

Determine the average velocity for the time interval $t=0$ to $t=2.0\text{ s}$

  $\begin{array}{c}{\overline{v}}_2=\frac{x_2-x_0}{\Delta t}\\ =\frac{1.71\text{ cm}-0}{2.0\text{ s}-0}\\ =0.86\text{ cm/s}\end{array}$

Find the position of the particle at time $t=1.0\text{ s}$ .

  $\begin{array}{c}{x}_1=\left(5.0\text{ cm}\right)\left[\sin \left(0.175{\text{ s}}^{-1}\right)t\right]\\ =\left(5.0\text{ cm}\right)\left[\sin \left(0.175{\text{ s}}^{-1}\right)\left(1.0\text{ s}\right)\right]\\ =0.87\text{ cm}\end{array}$

Determine the average velocity for the time interval $t=0$ to $t=1.0\text{ s}$

  $\begin{array}{c}{\overline{v}}_1=\frac{x_1-x_0}{\Delta t}\\ =\frac{0.87\text{ cm}-0}{1.0\text{ s}-0}\\ =0.87\text{ cm/s}\end{array}$

Find the position of the particle at time $t=0.5\text{ s}$ .

  $\begin{array}{c}{x}_{0.5}=\left(5.0\text{ cm}\right)\left[\sin \left(0.175{\text{ s}}^{-1}\right)t\right]\\ =\left(5.0\text{ cm}\right)\left[\sin \left(0.175{\text{ s}}^{-1}\right)\left(0.5\text{ s}\right)\right]\\ =0.437\text{ cm}\end{array}$

Determine the average velocity for the time interval $t=0$ to $t=0.5\text{ s}$

  $\begin{array}{c}{\overline{v}}_{0.5}=\frac{x_{0.5}-x_0}{\Delta t}\\ =\frac{0.437\text{ cm}-0}{0.5\text{ s}-0}\\ =0.87\text{ cm/s}\end{array}$

Find the position of the particle at time $t=0.25\text{ s}$ .

  $\begin{array}{c}{x}_{0.25}=\left(5.0\text{ cm}\right)\left[\sin \left(0.175{\text{ s}}^{-1}\right)t\right]\\ =\left(5.0\text{ cm}\right)\left[\sin \left(0.175{\text{ s}}^{-1}\right)\left(0.25\text{ s}\right)\right]\\ =0.22\text{ cm}\end{array}$

Determine the average velocity for the time interval $t=0$ to $t=0.25\text{ s}$

  $\begin{array}{c}{\overline{v}}_{0.25}=\frac{x_{0.25}-x_0}{\Delta t}\\ =\frac{0.22\text{ cm}-0}{0.25\text{ s}-0}\\ =0.87\text{ cm/s}\end{array}$

Conclusion:

Thus, the average velocities for the time intervals starting at $t=0$ and ending at $t=6.0,3.0,2.0,1.0,0.50\&0.25\text{ s}$ are $0.72\text{ cm/s, 0}\text{.86 cm/s, 0}\text{.86 cm/s, 0}\text{.87 cm/s, 0}\text{.87 cm/s and 0}\text{.87 cm/s}$ respectively.

(d)

To determine

To compute: $\frac{dx}{dt}$ at time $t=0$ and hence find the velocity at time $t=0$ .

Answer to Problem 114P

The value of $\frac{dx}{dt}$ and hence the velocity at time $t=0$ is found to be $\text{0}\text{.875 cm/s}$ .

Explanation of Solution

Given:

The equation for the position of the oscillating particle

  $x=\left(5.0\text{ cm}\right)\sin \left(0.175{\text{ s}}^{-1}\right)t$

Formula used:

The velocity of a particle is the first derivative of the position with respect to time and is given by,

  $v=\frac{dx}{dt}$

Calculation:

Differentiate the given equation with respect to time.

  $\begin{array}{c}v=\frac{dx}{dt}\\ =\frac{d}{dt}\left\{\left(5.0\text{ cm}\right)\left[\sin \left(0.175{\text{ s}}^{-1}\right)t\right]\right\}\\ =\left(5.0\text{ cm}\right)\left(0.175{\text{ s}}^{-1}\right)\left[\cos \left(0.175{\text{ s}}^{-1}\right)t\right]\\ =\left(0.875\text{ cm/s}\right)\left[\cos \left(0.175{\text{ s}}^{-1}\right)t\right]\end{array}$

Substitute $t=0$ in the above equation and calculate the velocity of the particle at time $t=0$ .

  $\begin{array}{c}v=\left(0.875\text{ cm/s}\right)\left[\cos \left(0.175{\text{ s}}^{-1}\right)t\right]\\ v_0=\left(0.875\text{ cm/s}\right)\left[\cos \left(0.175{\text{ s}}^{-1}\right)\left(0\right)\right]\\ =0.875\text{ cm/s}\end{array}$

Conclusion:

The value of $\frac{dx}{dt}$ and hence the velocity at time $t=0$ is found to be $\text{0}\text{.875 cm/s}$ .

(e)

To determine

To compare: the results of parts (c) and (d) and explain why the part(c) results approach part (d) result.

Explanation of Solution

Given:

Results of part (c)

The average velocities of the particle for the time intervals starting at $t=0$ and ending at $t=6.0,3.0,2.0,1.0,0.50\&0.25\text{ s}$

are as follows:

Time interval(s)	Average velocity (cm/s)
0-6.0	0.72
0-3.0	0.86
0-2.0	0.86
0-1.0	0.87
0-0.50	0.87
0.25	0.87

Results of part (d)

The instantaneous velocity of the particle at time $t=0$

  $v_0=0.875\text{ cm/s}$

Introduction:

Average velocity is defined as the ratio of change in position to the time interval.

  $\overline{v}=\frac{\Delta x}{\Delta t}$

The instantaneous velocity is given by,

  $v=\underset{\Delta t\to 0}{\lim }\frac{\Delta x}{\Delta t}$

As the measured time interval becomes smaller, the average velocity approaches the instantaneous velocity. For a large time interval such as $\Delta t=6.0$ , the value of the average velocity of $0.72\text{ cm/s}$ ,is not very close to the instantaneous velocity at $t=0$ of $0.875\text{ cm/s}$ . But as the time interval is shortened, the value comes very close to the value of the instantaneous velocity.

Conclusion:

Thus, it can be seen that as th magnitude of the measured time intervals decrease, the values of the average velocities approach the value of instantaneous velocity.