Chapter 11, Problem 60P

To determine

To Find: The number of different standing wave patterns achieved by varying.

Answer to Problem 60P

Four different standing wave patterns are achieved by varying L .

Explanation of Solution

Given:

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

Hanging mass, $m=0.080\text{kg}$

Acceleration due to gravity, $g=9.8{\text{m/s}}^{\text{2}}$

Concept used:

The tension in the string is the weight of hanging mass,

  $F_T=mg$

And the speed of the wave on the string is

  $v=\sqrt{\frac{F_T}{\mu }}$

Thus, the speed of the wave on the string is,

  $v=\sqrt{\frac{\text{mg}}{\mu }}$

Write the expression for the wavelength.

  $\begin{array}{c}\lambda =\frac{v}{f}\\ \lambda =\frac{1}{f}\sqrt{\frac{\text{mg}}{\mu }}\end{array}$

Here, $f$ is the frequency of the wave,

  $m$ is the hanging mass,

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

Calculation:

From the above equation

  $\lambda =\frac{1}{60\text{Hz}}\sqrt{\frac{0.080\text{kg}\times 9.8{\text{m/s}}^{\text{2}}}{3.9\times {10}^{-4}\text{kg/m}}}$

  $\lambda =0.747\text{m}$

The length of the string must be the integer of multiple halves of the wavelength So,

  $L=\frac{\lambda }{2},\lambda ,\frac{3\lambda }{2},2\lambda ,\mathrm{....}$

Substituting the values,

  $\begin{array}{c}\frac{\lambda }{2}=\frac{0.747}{2}=0.3735\\ \lambda =0.747\\ \frac{3\lambda }{2}=\frac{0.747}{2}\times 3\\ =1.1205\\ 2\lambda =0.747\times 2\\ =1.494\end{array}$

There are four standing wave patterns achieved.

Conclusion:

Therefore, total wave patterns achieved are 4.