Chapter 15, Problem 2SP

(a)

To determine

What is the mass per unit length of the string?

Answer to Problem 2SP

The mass per unit length of the string is $3.9\times {10}^{-2}\text{kg}/\text{m}$.

Explanation of Solution

Given info: length of guitar string is $1.28\text{m}$ , mass of the guitar is $0.05\text{kg}$.

Write the equation to find the mass per unit length of the guitar.

$m=\frac{M}{L}$

Here,

$m$ is the mass per unit length of the string

$M$ is the mass

$L$ is the length of the string

Substitute $0.05\text{kg}$ for $M$ and $1.28\text{m}$ for $L$ in the equation to get $m$.

$\begin{array}{c}m=\frac{0.05\text{kg}}{1.28\text{m}}\\ =3.9\times {10}^{-2}\text{kg}/\text{m}\end{array}$

Conclusion:

Therefore, the mass per unit length of the string is $3.9\times {10}^{-2}\text{kg}/\text{m}$.

(b)

To determine

The speed of waves in the tightened string.

Answer to Problem 2SP

The speed of waves in the tightened string is $144\text{m}/\text{s}$.

Explanation of Solution

Given info: The tension in the guitar string is $810\text{N}$.

Write the equation to find the speed of wave in the stretched string.

$v=\sqrt{\frac{F}{m}}$

Here,

$v$ is the speed of waves in the stretched string

$F$ is the tension in the string

$m$ is the mass per unit length of the string

Substitute $810\text{N}$ for $F$ and $3.9\times {10}^{-2}\text{kg}/\text{m}$ for $m$ in the equation to get $v$.

$\begin{array}{c}v=\sqrt{\frac{810\text{N}}{3.9\times {10}^{-2}\text{kg}/\text{m}}}\\ =144\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed of waves in the stretched string is $144\text{m}/\text{s}$.

(c)

To determine

The wavelength of waves that forms fundamental standing waves.

Answer to Problem 2SP

The wavelength of the waves formed in the string that forms fundamental standing wave is $1.42\text{m}$.

Explanation of Solution

Given info: length of string from one end to other end is $71\text{cm}$.

A fundamental standing wave is formed by interference of the waves in the medium. A fundamental standing wave has nodes at either end of the string. The wavelength of the wave formed here is double the length of the string through which it passes.

Write the equation to find the wavelength of fundamental standing wave formed in the string.

$\lambda =2L$

Here,

$\lambda$ is the wavelength of fundamental wavelength of the standing wave

$L$ is the length of the string

Substitute $71\text{cm}$ for $L$ in the equation to get $\lambda$.

$\begin{array}{c}\lambda =2\times 71\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =1.42\text{m}\end{array}$

Conclusion:

Therefore, the wavelength of the fundamental standing wave formed is $1.42\text{m}$.

(d)

To determine

The frequency of the fundamental wave.

Answer to Problem 2SP

The frequency of the fundamental wave is $101\text{Hz}$.

Explanation of Solution

Write the equation to find the frequency of the fundamental wave.

$f=\frac{v}{\lambda }$

Here,

$f$ is the frequency of fundamental wave

$v$ is the speed of fundamental wave

$\lambda$ is the wavelength of the fundamental wave

Substitute $1.42\text{m}$ for $\lambda$ and $144\text{m}/\text{s}$ for $v$ in the equation to find $f$.

$\begin{array}{c}f=\frac{144\text{m}/\text{s}}{1.42\text{m}}\\ =101\text{Hz}\end{array}$

Conclusion:

Therefore, the frequency of the fundamental wave is $101\text{Hz}$.

(e)

To determine

The frequency and wavelength of next harmonic.

Answer to Problem 2SP

The frequency of waves in the second harmonic is $203\text{Hz}$ and the wavelength of second wavelength is $0.71\text{m}$.

Explanation of Solution

Write the equation to find the wavelength of the wave in second harmonics.

$\lambda =\frac{2L}{n}$

Here,

$\lambda$ is the wavelength of the wave in second harmonic

$L$ is the length of string

$n$ is the order of harmonic

Substitute $71\text{cm}$ for $L$ and $2$ for $n$ in the equation to get $\lambda$.

$\begin{array}{c}\lambda =\frac{2\times 71\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)}{2}\\ =0.71\text{m}\end{array}$

Similarly write the equation to find the frequency of wave.

$f=\frac{v}{\lambda }$

Here,

$f$ is the frequency of the wave

$v$ is the velocity of wave

$\lambda$ is the wavelength of the wave

Substitute $144\text{m}/\text{s}$ for $v$ and $0.71\text{m}$ for $\lambda$ in the equation to get $f$.

$\begin{array}{c}f=\frac{144\text{m}/\text{s}}{0.71\text{m}}\\ =203\text{Hz}\end{array}$

Conclusion:

Therefore, the frequency of waves in the second harmonic is $203\text{Hz}$ and the wavelength of second wavelength is $0.71\text{m}$.