Chapter 15, Problem 15.76AP

When a block of mass M, connected to the end of a spring of mass m_s = 7.40 g and force constant k, is set into simple harmonic motion, the period of its motion is

$T=2\pi \sqrt{\frac{M+\left(m_s/3\right)}{k}}$

A two-part experiment is conducted with the use of blocks of various masses suspended vertically from the spring as shown in Figure P15.76. (a) Static extensions of 17.0, 29.3, 35.3, 41.3, 47.1, and 49.3 cm are measured for M values of 20.0, 40.0, 50.0, 60.0, 70.0, and 80.0 g, respectively. Construct a graph of Mg versus x and perform a linear least-squares fit to the data. (b) From the slope of your graph, determine a value for k for this spring. (c) The system is now set into simple harmonic motion, and periods are measured with a stopwatch. With M = 80.0 g, the total time interval required for ten oscillations is measured to be 13.41 s. The experiment is repeated with M values of 70.0, 60.0, 50.0, 40.0, and 20.0 g, with corresponding time intervals for ten oscillations of 12.52, 11.67, 10.67, 9.62, and 7.03 s. Make a table of these masses and times. (d) Compute the experimental value for T from each of these measurements. (e) Plot a graph of T² versus M and (f) determine a value for k from the slope of the linear least-squares fit through the data points. (g) Compare this value of k with that obtained in part (b). (h) Obtain a value for m_s from your graph and compare it with the given value of 7.40 g.

To determine

To draw: The graph of $Mg$ versus $x$ and perform the linear least- square fit to the data.

Answer to Problem 15.76AP

The graph of $Mg$ versus $x$ is,

Explanation of Solution

Given info: The block of mass is $M$, the mass of spring is $7.40\text{g}$ and the force constant is $k$.

The values of mass and the static extension are given and calculate the value of the $Mg$.

Static extension (x) in $\text{m}$	Mass (M) in $\text{kg}$	Weight $\left(Mg\right)$ in $\text{N}$
0.17	0.02	0.196
0.293	0.04	0.392
0.353	0.05	0.49
0.413	0.06	0.588
0.471	0.07	0.686
0.493	0.08	0.784

Table (1)

Conclusion:

The table (1) indicates the values required to plot the graph of $Mg$ versus $x$.

Figure (1)

To determine

The value of $k$ for spring using the slope of graph.

Answer to Problem 15.76AP

The value of $k$ for spring using the slope of graph is $1.7386\text{N}/\text{m}$.

Explanation of Solution

Given info: The block of mass is $M$, the mass of spring is $7.40\text{g}$ and the force constant is $k$.

The equation of the graph is,

$y=1.7386x-0.1128$

The slope intercept form of the equation of the line is,

$y=mx+c$

Here,

$m$ is the slope.

$c$ is the intercept.

Compare equation (1) and (2).

$m=1.7386$

Since the slope of the graph indicates the force constant of the spring.

$k=1.7386\text{N}/\text{m}$

Conclusion:

Therefore, the value of $k$ for spring using the slope of graph is $1.7386\text{N}/\text{m}$.

To determine

To draw: The table of given masses and the times.

Answer to Problem 15.76AP

The table of given masses and the times is,

Mass in $\text{g}$	Time intervals in $\text{s}$.
20	7.03
40	9.62
50	10.67
60	11.67
70	12.52
80	13.41

Explanation of Solution

Given info: The block of mass is $M$, the mass of spring is $7.40\text{g}$ and the force constant is $k$.

The values of mass and the static extension are given and calculate the value of the $Mg$.

Mass in $\text{g}$	Time intervals in $\text{s}$.
20	7.03
40	9.62
50	10.67
60	11.67
70	12.52
80	13.41

Table (2)

Conclusion:

The table (1) indicates the values of mass and the time intervals.

To determine

The experimental value for $T$ from each of the measurements.

Answer to Problem 15.76AP

The experimental values for $T$ from each of the measurements are $0.703\text{s}$, $0.962\text{s}$, $1.067\text{s}$, $1.167\text{s}$, $1.252\text{s}$ and $1.341\text{s}$.

Explanation of Solution

Given info: The block of mass is $M$, the mass of spring is $7.40\text{g}$ and the force constant is $k$.

The expression for the time periods for each experiment is,

$T_{\text{n}}=\frac{T}{\text{n}}$

Here,

$n$ is total number of experiments.

The ten experiments are conducted.

Calculate time periods for each experiment.

Total time period $\left(T\right)$ in $\text{s}$	Time period for one experiment $\left(T_{\text{n}}=\frac{T}{\text{10}}\text{s}\right)$
7.03	0.703
9.62	0.962
10.67	0.1067
11.67	1.167
12.52	1.252
13.41	1.341

Table (3)

Conclusion:

Therefore, the experimental values for $T$ from each of the measurements are $0.703\text{s}$, $0.962\text{s}$, $1.067\text{s}$, $1.167\text{s}$, $1.252\text{s}$ and $1.341\text{s}$.

To determine

To draw: The graph of $T^2$ versus $M$.

Answer to Problem 15.76AP

The graph of $T^2$ versus $M$ is,

Explanation of Solution

Given info: The block of mass is $M$, the mass of spring is $7.40\text{g}$ and the force constant is $k$.

Time period for one experiment $\left(T\right)$ in $\text{s}$	$T^2$	Mass (M) in $\text{kg}$
0.703	0.494209	0.02
0.962	0.925444	0.04
0.1067	1.138489	0.05
1.167	1.361889	0.06
1.252	1.567504	0.07
1.341	1.798281	0.08

Table (4)

The table (4) indicates the values required to plot the graph of $T^2$ versus $M$.

Figure (2)

To determine

The value of $k$ for spring using the slope of graph.

Answer to Problem 15.76AP

The value of $k$ for spring using the slope of graph is $1.82\text{N}/\text{m}$.

Explanation of Solution

Given info: The block of mass is $M$, the mass of spring is $7.40\text{g}$ and the force constant is $k$.

The equation of the graph is,

$\begin{array}{c}y=21.665x+0.0589\\ T^2=21.665M+0.0589\end{array}$ (3)

The given expression is,

$T=2\pi \sqrt{\frac{M+\frac{m_{\text{s}}}{3}}{k}}$

Square both sides in above expression.

$T^2=\frac{4{\pi }^2}{k}M+\frac{4{\pi }^2}{3k}{m}_{\text{s}}$ (4)

Compare equation (3) and (4).

$\begin{array}{c}\frac{4{\pi }^2}{k}=21.665\\ k=1.82\text{N}/\text{m}\end{array}$

Conclusion:

Therefore, the value of $k$ for spring using the slope of graph is $1.82\text{N}/\text{m}$.

To determine

The comparison in value of $k$ obtained between the part (b) and part (f) of the question.

Answer to Problem 15.76AP

The value of $k$ obtained in part (b) of the question is less than the value of $k$ obtained in part (f) of the question.

Explanation of Solution

Given info: The block of mass is $M$, the mass of spring is $7.40\text{g}$ and the force constant is $k$.

The value of $k$ obtained in part (b) of the question is,

$k_{\text{b}}=1.7386\text{N}/\text{m}$

The value of $k$ obtained in part (f) of the question is,

$k_{\text{f}}=1.82\text{N}/\text{m}$

Compare the values.

$\begin{array}{c}{k}_{\text{b}}<k_{\text{f}}\\ 1.7386\text{N}/\text{m}<1.82\text{N}/\text{m}\end{array}$

Conclusion:

Therefore, the value of $k$ obtained in part (b) of the question is less than the value of $k$ obtained in part (f) of the question.

To determine

The value of $m_{\text{s}}$ for spring using graph and compare it with $7.40\text{g}$

Answer to Problem 15.76AP

The value of $m_{\text{s}}$ for spring using graph is $8.14\text{g}$ and it is greater than $7.40\text{g}$.

Explanation of Solution

Given info: The block of mass is $M$, the mass of spring is $7.40\text{g}$ and the force constant is $k$.

The equation of the graph is,

$\begin{array}{c}y=21.665x+0.0589\\ T^2=21.665M+0.0589\end{array}$

The given expression is,

$T=2\pi \sqrt{\frac{M+\frac{m_{\text{s}}}{3}}{k}}$

Square both sides in above expression.

$T^2=\frac{4{\pi }^2}{k}M+\frac{4{\pi }^2}{3k}{m}_{\text{s}}$

Compare both the above expression.

$\frac{4{\pi }^2}{3k}{m}_{\text{s}}=0.0589$

Substitute $1.82\text{N}/\text{m}$ for $k$ in above expression.

$\begin{array}{c}\frac{4{\pi }^2}{3\times 1.82\text{N}/\text{m}}{m}_{\text{s}}=0.0589\\ m_{\text{s}}=8.14\times {10}^{-3}\text{kg}\times \frac{1000\text{g}}{1\text{kg}}\\ =8.14\text{g}\end{array}$

Conclusion:

Therefore, the value of $m_{\text{s}}$ for spring using graph is $8.14\text{g}$ and it is greater than $7.40\text{g}$.