Chapter 15, Problem 2SP

A guitar string has an overall length of 1.25 m and a total mass of 1.0 g (0.001 kg) before it is strung on the guitar. Once on the guitar, however, there is a distance of 69 cm between its fixed end points. It is tightened to a tension of 74 N.

a.    What is the mass per unit of length of this string?

b.    What is the wave speed for waves on the tightened string?

c.    What is the wavelength of the traveling waves that interfere to form the fundamental standing wave (nodes just at either end) for this string?

d.    What is the frequency of the fundamental wave?

e.    What are the wavelength and frequency of the next harmonic (with a node in the middle of the string)?

To determine

The mass per unit length of the string.

Answer to Problem 2SP

The mass per unit length of the string is $8.0\times {10}^{-4}\text{kg}/\text{m}$.

Explanation of Solution

Given Info: The length of the string is $1.25\text{m}$ and mass of the string is $0.001\text{kg}$.

Write the expression to calculate the mass per unit length of the string.

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

Here,

$\mu$ is the mass per unit length of the string

L is the length of the string

m is the mass of the string

Substitute $1.25\text{m}$ for L and $0.001\text{kg}$ for m in the above equation to calculate $\mu$.

$\begin{array}{c}\mu =\frac{0.001\text{kg}}{1.25\text{m}}\\ =0.0008\text{kg}/\text{m}\\ =8.0\times {10}^{-4}\text{kg}/\text{m}\end{array}$

Conclusion:

Therefore, the mass per unit length of the string is $8.0\times {10}^{-4}\text{kg}/\text{m}$.

To determine

The wave speed.

Answer to Problem 2SP

The wave speed is $304\text{m}/\text{s}$.

Explanation of Solution

Given Info: The mass per unit length of the string is $0.0008\text{kg}/\text{m}$ and tension on the string is $74\text{N}$.

Write the formula to calculate the wave speed.

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

Here,

v is the wave speed

F is the tension on the string

Substitute $0.0008\text{kg}/\text{m}$ for $\mu$ and $74\text{N}$ for F in the above equation to calculate v.

$\begin{array}{c}v=\sqrt{\frac{74\text{N}}{0.0008\text{kg}/\text{m}}}\\ \approx 304\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the wave speed is $304\text{m}/\text{s}$.

To determine

The wavelength of the travelling wave.

Answer to Problem 2SP

The wavelength is $1.38\text{m}$.

Explanation of Solution

Given Info: The distance between two end is $69\text{cm}$.

Write the expression to calculate the wavelength.

$\frac{\lambda }{2}=l$

Here,

$\lambda$ is the wavelength

l is the distance between two ends

Substitute $69\text{cm}$ for l in the above equation to calculate $\lambda$.

$\begin{array}{c}\frac{\lambda }{2}=69\text{cm}\\ \lambda =69\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)2\\ =1.38\text{m}\end{array}$

Conclusion:

Therefore, the wavelength is $1.38\text{m}$.

To determine

The frequency of the standing wave.

Answer to Problem 2SP

The frequency of the standing wave is $220\text{Hz}$.

Explanation of Solution

Given Info: The wavelength of the standing wave is The longest possible wavelength is $1.38\text{m}$ and speed is $304\text{m}/\text{s}$.

Write the expression to calculate the speed of the sound wave.

$v=f\lambda$

Here,

$\lambda$ is the wavelength of the standing wave which is longest possible wavelength

f is the frequency of the sound wave

Substitute $304\text{m}/\text{s}$ for v and $1.38\text{m}$ for $\lambda$ in the above equation to calculate f.

$\begin{array}{c}f\left(1.38\text{m}\right)=304\text{m}/\text{s}\\ f=\frac{304\text{m}/\text{s}}{1.38\text{m}}\\ \approx 220\text{Hz}\end{array}$

Conclusion:

Therefore, the frequency of the standing wave is $220\text{Hz}$.

To determine

The frequency and wavelength of the next harmonic wave.

Answer to Problem 2SP

The frequency of the second harmonic wave is $440\text{Hz}$ and wave length is $0.69\text{m}$.

Explanation of Solution

Given Info: The distance between two end is $69\text{cm}$ and speed is $304\text{m}/\text{s}$.

Write the formula to calculate the wavelength of the second harmonic.

${\lambda }^{\prime }=l$

Here,

${\lambda }^{\prime }$ is the second harmonic wavelength.

Substitute $69\text{cm}$ for l in the above equation to calculate ${\lambda }^{\prime }$.

$\begin{array}{c}{\lambda }^{\prime }=69\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =0.69\text{m}\end{array}$

Write the expression to calculate the speed of the sound wave.

$v=f^{\prime }{\lambda }^{\prime }$

Here,

$f^{\prime }$ is the frequency of the second harmonic

Substitute $304\text{m}/\text{s}$ for v and $0.69\text{m}$ for ${\lambda }^{\prime }$ in the above equation to calculate $f^{\prime }$.

$\begin{array}{c}{f}^{\prime }\left(0.69\text{m}\right)=304\text{m}/\text{s}\\ f^{\prime }=\frac{304\text{m}/\text{s}}{0.69\text{m}}\\ \approx 440\text{Hz}\end{array}$

Conclusion:

Therefore, the frequency of the second harmonic wave is $440\text{Hz}$ and wave length is $0.69\text{m}$.