Standing Waves and Resonance — Question 7

Question 7

A small bead shifts an antinode-loaded mode more than a node-loaded mode

A fixed-end string has total mass $M=\mu L$. A small bead of mass $m_b\ll M$ is attached at $x=a$. For an approximately unchanged mode shape $\varphi_n=\sin(n\pi x/L)$, use effective mass $M/2+m_b\varphi_n(a)^2$, with unchanged modal stiffness.

1. Find the approximate frequency ratio after attaching the bead.

2. For $m_b=.05M$ and $a=L/2$, estimate the fundamental shift and compare with mode $n=2$. State the limitation of the estimate.

Question 7 - Solution

A small bead shifts an antinode-loaded mode more than a node-loaded mode

Method and mathematical working

Keep the unperturbed mode shape as the stated approximation. Add the bead contribution to effective modal mass while retaining the original modal stiffness.$$\begin{aligned}M_{\mathrm{eff}}&=M/2+m_b\sin^2(n\pi a/L),\\\frac{\omega_n^{\prime}}{\omega_n}&=\sqrt{\frac{M/2}{M/2+m_b\sin^2(n\pi a/L)}}.\\[4pt]\frac{f_1^{\prime}}{f_1}&=(1+2\cdot0.05)^{-1/2}=0.95346,\\\sin(2\pi a/L)\big|_{a=L/2}&=0\Rightarrow f_2^{\prime}/f_2=1.\end{aligned}$$

(a) Calculation and conclusion

Since frequency is proportional to inverse square root of effective mass, $f_n^{\prime}/f_n\approx[1+(2m_b/M)\sin^2(n\pi a/L)]^{-1/2}$. The bead contributes according to the squared local mode displacement.

(b) Interpretation and limits

The fundamental has a midpoint antinode, giving $1/\sqrt{1.10}=.95346$, about a $4.65\%$ decrease. Mode 2 has a midpoint node and is unchanged in the ideal point-bead model for that exact shape. The estimate for affected modes neglects changes in shape and is most reliable for small added mass.

Physical check

At a node the bead has no transverse motion in this approximation and adds no modal kinetic energy. At an antinode it gives the largest frequency reduction. Large bead masses would invalidate the unchanged-shape assumption.