Chapter 11, Problem 59P

a.

To determine

To find: The mass m hangs from the end of the string.

Answer to Problem 59P

The mass m hang from the end of string is $1.289\text{kg}$ .

Explanation of Solution

Given:

length $\mathrm{l}=1.50\text{m}$

Mass per unit length $\mathrm{\mu }=3.9\times {10}^{-4}\text{kg/m}$

Frequency $\mathrm{f}=60\text{Hz}$

No. of loops $\mathrm{n}=1$

Concept used:

Write the expression for the mass m hangs from the end of string.

  $\mathrm{m}=\frac{4{\mathrm{l}}^2{\mathrm{f}}^2\mathrm{\mu }}{{\mathrm{n}}^2\mathrm{g}}$

Here, $\mathrm{l}$ is the length of the string,

f is frequency

  $\mathrm{n}$ is the number of loops,

  $\mathrm{\mu }$ is the mass per unit length,

  $\mathrm{g}$ is the acceleration due to gravity.

Calculation:

From the above equation

  $\begin{array}{l}\mathrm{m}=\frac{4{(1.50)}^2{\left(60\right)}^2(3.9\times {10}^{-4}\text{kg/m)}}{{(1)}^2(9.8\times {\text{m/s}}^{\text{2}})}\\ \mathrm{m}=1.289\text{kg}\end{array}$

Conclusion:

The mass m hang from the end of string is $1.289\text{kg}$.

b.

To determine

To find: The mass m hangs from the end of the string.

Answer to Problem 59P

The mass hang from the end of string is $0.322\text{kg}$ .

Explanation of Solution

Given:

length $\mathrm{l}=1.50\text{m}$

Mass per unit length $\mathrm{\mu }=3.9\times {10}^{-4}\text{kg/m}$

Frequency $\mathrm{f}=60\text{Hz}$

No. of loops $\mathrm{n}=2$

Concept used:

Write the expression for the mass m hangs from the end of string.

  $\mathrm{m}=\frac{4{\mathrm{l}}^2{\mathrm{f}}^2\mathrm{\mu }}{{\mathrm{n}}^2\mathrm{g}}$

Here, $\mathrm{l}$ is the length of the string,

  $\mathrm{n}$ is the number of loops,

f is frequency

  $\mathrm{\mu }$ is the mass per unit length,

  $\mathrm{g}$ is the acceleration due to gravity.

Calculation:

From the above equation

  $\begin{array}{l}\mathrm{m}=\frac{4{(1.50)}^2{\left(60\right)}^2(3.9\times {10}^{-4}\text{kg/m)}}{{(2)}^2(9.8\times {\text{m/s}}^{\text{2}})}\\ \mathrm{m}=0.322\text{kg}\end{array}$

Conclusion:

Mass hang from the end of the string is $0.322\text{kg}$ .

c.

To determine

To find: The mass m hangs from the end of the string.

Answer to Problem 59P

The mass m hang from the end of string is $0.0515\text{kg}$ .

Explanation of Solution

Given:

length $\mathrm{l}=1.50\text{m}$

Mass per unit length $\mathrm{\mu }=3.9\times {10}^{-4}\text{kg/m}$

Frequency $\mathrm{f}=60\text{Hz}$

No. of loops $\mathrm{n}=5$

Concept used:

Write the expression for the mass m hangs from the end of string.

  $\mathrm{m}=\frac{4{\mathrm{l}}^2{\mathrm{f}}^2\mathrm{\mu }}{{\mathrm{n}}^2\mathrm{g}}$

Here, $\mathrm{l}$ is the length of the string,

  $\mathrm{n}$ is the number of loops,

f is frequency

  $\mathrm{\mu }$ is the mass per unit length,

  $\mathrm{g}$ is the acceleration due to gravity.

Calculation:

From the above equation

  $\begin{array}{l}\mathrm{m}=\frac{4{(1.50)}^2{\left(60\right)}^2(3.9\times {10}^{-4}\text{kg/m)}}{{(5)}^2(9.8\times {\text{m/s}}^{\text{2}})}\\ \mathrm{m}=0.0515\text{kg}\end{array}$

Conclusion:

Mass hang from the end of the string is $0.0515\text{kg}$ .