RES.8-005 | Fall 2012 | Undergraduate

# Vibrations and Waves Problem Solving

Standing Waves Part II

## Problems

### Standing Waves Part II

#### Problem 1

An ideal taut string of length $$l$$ and mass $$m$$ is attached to a fixed point at one end and to a massive ring of mass $$M_R$$ at the other end as shown below. The ring is free to move on a horizontal frictionless rod which is perpendicular to the string in its equiibrium position. The tension in the string, $$T$$, can be considered constant at all times and gravity can be ignored.

1. Starting with the general form of the normal mode solutions of the classical wave equation for a taut string, derive an equation which, when solved, gives the angular frequency $$\omega$$ of the lowest mode of the system. YOU ARE NOT ASKED TO SOLVE THIS EQUATION.
2. Determine the angular frequency of the lowest normal mode for
• $$M_R = 0$$
• $$M_R = \infty$$.

1. \begin{eqnarray} \nonumber \omega &=& \frac{T}{M_R}\sqrt{\frac{m}{Tl}}\cot \left( \omega l \sqrt{\frac{m}{Tl}}\right) \\ \nonumber &=& \frac{1}{M_R}\sqrt{\frac{Tm}{l}}\cot \left( \omega \sqrt{\frac{ml}{T}}\right) \end{eqnarray}

This is a transcendental equation with no simple solution.

• For $$M_R = 0$$, we get \begin{eqnarray} \nonumber &&\cot \left( \frac{\omega}{v}l\right) = 0 \hspace{3mm} with \hspace{3mm} v = \sqrt{\frac{Tl}{m}} \\ \nonumber &\Rightarrow& \frac{\omega}{v}l = \frac{\pi}{2} \\ \nonumber &\Rightarrow& \omega = \frac{\pi}{2l}v = \frac{\pi}{2l}\sqrt{\frac{Tl}{m}} = \frac{\pi}{2}\sqrt{\frac{T}{ml}} \end{eqnarray} for the lowest normal mode. Intuitively, this makes sense since this situation corresponds to a string which is fixed at one end and free at the other, so the lowest normal mode has $$l = \lambda / 4$$ or $$\lambda = 4l$$ and thus $$\omega = \dfrac{2 \pi v}{\lambda} = \dfrac{2 \pi v}{4l} = \dfrac{\pi}{2l}v$$.
• For $$M_R = \infty$$, we get \begin{eqnarray} \nonumber &&\tan \left( \frac{\omega}{v}l\right) = 0 \\ \nonumber &\Rightarrow& \frac{\omega}{v}l = \pi \\ \nonumber &\Rightarrow& \omega = \frac{\pi}{l}v = \frac{\pi}{l}\sqrt{\frac{Tl}{m}} = \pi\sqrt{\frac{T}{ml}} \end{eqnarray} for the lowest normal mode. Intuitively, this makes sense since this situation corresponds to a string which is fixed at both ends, so the lowest normal mode has $$l = \lambda / 2$$ or $$\lambda = 2l$$ and thus $$\omega = \dfrac{2 \pi v}{\lambda} = \dfrac{2 \pi v}{2l} = \dfrac{\pi}{l}v$$.

#### Problem 2

A system consists of two materials glued together at $$x = L$$ and subject to longitudinal (sound) wave propagation. The end at $$x = 0$$ is fixed and the end at $$x = 4L$$ is free. A tungsten-lead alloy is used for the segment $$0 \leq x \leq L$$, with a Young’s modulus of $$Y = 7 \times 10^{10}$$ N/m$$^2$$ and density $$\rho = 14.4 \times 10^{3}$$ kg/m$$^3$$, while the segment $$L \leq x \leq 4L$$ is comprised of a carbon fiber alloy that is as strong as the tungsten-lead alloy with $$Y = 7 \times 10^{10}$$ N/m$$^2$$ but nine times less dense, with $$\rho = 1.6 \times 10^{3}$$ kg/m$$^3$$.

1. Derive a condition for the normal mode frequency of longitudinal oscillation, $$\omega_n$$, of the two-metal system.
2. Evaluate the frequencies for the first four modes, in terms of the length $$L$$.

View/Hide Hint

The phase velocity $$v$$ of longitudinal waves $$= \sqrt{\dfrac{Y}{\rho}}$$
$$\therefore \dfrac{v_2}{v_1} = 3$$ and the essence of the problem is the boundary conditions at $$x$$ = 0, $$L$$ and $$4L$$.

1. $$sin{\dfrac{\omega_n L}{v_1}} = \pm \dfrac{\sqrt{3}}{2}$$      with $$v_1 = \sqrt{\dfrac{Y_1}{\rho_1}} \approx 2.2 \times 10^{3}$$ m/s
2. $$\omega_1 = {\dfrac{v_1}{L}}{\dfrac{\pi}{3}}$$;      $$\omega_2 = {\dfrac{v_1}{L}}{\dfrac{2 \pi}{3}}$$;      $$\omega_3 = {\dfrac{v_1}{L}}{\dfrac{4 \pi}{3}}$$;      $$\omega_4 = {\dfrac{v_1}{L}}{\dfrac{5 \pi}{3}}$$

[Note that just about any factor other than 3 for speed and length ratios would give an equation which could not be easily solved.]

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Fall 2012
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