Chapter 12, Problem 77GP

To determine

To Find: The tension in the string.

Answer to Problem 77GP

  $146.4\text{ N}$

Explanation of Solution

Given:

Mass of guitar string, $\mathrm{m}=\text{2}.\text{1 g}=\text{ 2}.\text{1}\times \text{1}{0}^{-\text{3}}\text{kg}$

Length of string, $\mathrm{L} =\text{75 cm = 0}\text{.75 m}$

Tube opened at one end.

Formula used:

Write the expression for the speed of the sound.

  $\mathrm{v}=\sqrt{\frac{\mathrm{T}}{\mathrm{\mu }}}$

  $\mathrm{T}={\mathrm{v}}^2\mathrm{\mu }$

Here, $\mathrm{v}$ is the speed of sound.

T is the tension in the string.

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

Calculation:

Tension in the guitar string is determined by the following equation,

  $\mathrm{T}={\mathrm{v}}^2\mathrm{\mu }$

Now, determine the speed of the sound on the guitar ( v ),

The frequency of the string is

  $\mathrm{f}=\frac{\mathrm{v}}{\mathrm{L}}\text{Hz}$

  $\mathrm{f}=\frac{343}{0.75}\text{ Hz}$

  $\mathrm{f}=457.33\text{ Hz}$

  $\mathrm{f}={\mathrm{f}}_{\mathrm{s}}={\mathrm{f}}_3=457.33\text{ Hz}$

The fundamental standing frequency is equal to the frequency at the third harmonic tube.

Now, calculate the wavelength of third string is,

  ${\mathrm{\lambda }}_3=\frac{2\mathrm{L}}{\mathrm{n}}$

Where, n = third string = 3.

  ${\mathrm{\lambda }}_3=\frac{2\times 0.75}{3}$

  ${\mathrm{\lambda }}_3=0.5\text{ m}$

Now, calculate the speed of sound,

  $\mathrm{v}=\mathrm{f}\times {\mathrm{\lambda }}_3\text{ m/s}$

  $\mathrm{v}=457.33\times 0.5\text{ m/s}$

  $\mathrm{v}=228.66\text{ m/s}$

Substitute the values in the equation (tension) T

  $\mathrm{T}={\mathrm{v}}^2\mathrm{\mu }$

  $\mathrm{T}={(228.66\text{ m/s})}^2\times \left(\frac{2.1\times {\text{10}}^{-3}\text{kg}}{0.75\text{ m}}\right)$

  $\mathrm{T}=146.4\text{ N}$ .

Conclusion:

The tension that should be in the string is $146.4\text{ N}$ when it is to produce resonance with third harmonic in the tube.