A drumskin is stretched across a fixed circular rim of radius a. Small transverse vibrations of the skin have an amplitude z(ρ, Φ, t) that satisfies ∇²z = 1 ∂²z/c² ∂t² in plane polar coordinates. For a normal mode independent of azimuth, in which case z = Z(ρ) cos ωt, find the differential equation

Question

A drumskin is stretched across a fixed circular rim of radius a. Small transverse vibrations of the skin have an amplitude z(ρ, [latex]\phi[/latex], t) that satisfies

[latex] abla^2 z=\frac{1}{c^2} \frac{\partial^2 z}{\partial t^2}[/latex]

in plane polar coordinates. For a normal mode independent of azimuth, in which case z = Z(ρ) cos ωt, find the differential equation satisfied by Z(ρ). By using a trial function of the form [latex]a^ν- ho^ν[/latex], with adjustable parameter [latex]ν[/latex], obtain an estimate for the lowest normal mode frequency.

[ The exact answer is [latex](5.78)^{1 / 2}[/latex]c/a.]

Accepted Answer

In cylindrical polar coordinates, (ρ, [latex]\phi[/latex]), the wave equation,

[latex] abla^2 z=\frac{1}{c^2} \frac{\partial^2 z}{\partial t^2},[/latex]

has azimuth-independent solutions (i.e. independent of [latex]\phi[/latex]) of the form z(ρ, t) = Z(ρ) cos ωt, and reduces to

[latex]\begin{aligned}& \frac{1}{ ho} \frac{d}{d ho}\left( ho \frac{d Z}{d ho} ight) \cos \omega t=-\frac{Z \omega^2}{c^2} \cos \omega t, \& \frac{d}{d ho}\left( ho \frac{d Z}{d ho} ight)+\frac{\omega^2}{c^2} ho Z=0.\end{aligned}[/latex]

The boundary conditions require that Z(a) = 0 and, so that there is no physical discontinuity in the slope of the drumskin at the origin, [latex]Z^{\prime}(0) = 0[/latex].

This is an S–L equation with p = ρ, q = 0 and weight function w = ρ. A suitable trial function is [latex]Z( ho)=a^ν- ho^ν[/latex] , which automatically satisfies Z(a) = 0 and, provided [latex]ν > 1[/latex], has [latex]Z^{\prime}(0)=-\left.ν ho^{ν-1} ight|_{ ho=0}=0[/latex].

We recall that the lowest eigenfrequency satisfies the general formula

[latex]\frac{\omega^2}{c^2} \leq \frac{\int_0^a\left[\left(p Z^{\prime} ight)^2-q Z^2 ight] d ho}{\int_0^a w Z^2 d ho}.[/latex]

In this case

[latex]\begin{aligned}\frac{\omega^2}{c^2} & \leq \frac{\int_0^a ho ν^2 ho^{2 ν-2} d ho}{\int_0^a ho\left(a^ν- ho^ν ight)^2 d ho} \& =\frac{\int_0^a ν^2 ho^{2 ν-1} d ho}{\int_0^a\left( ho a^{2 ν}-2 ho^{ν+1} a^ν+ ho^{2 ν+1} ight) d ho} \& =\frac{\left(ν^2 a^{2 ν} ight) / 2 ν}{\frac{a^{2 ν+2}}{2}-\frac{2 a^{2 ν+2}}{ν+2}+\frac{a^{2 ν+2}}{2 ν+2}} \& =\frac{1}{a^2} \frac{ν(ν+2)(2 ν+2)}{(ν+2)(2 ν+2)-4(2 ν+2)+2(ν+2)} \& =\frac{(ν+2)(ν+1)}{ν a^2}\end{aligned}[/latex]

Since [latex]ν[/latex] is an adjustable parameter and we know that, however we choose it, the resulting estimate can never be less than the lowest true eigenvalue, we choose the value that minimises the above estimate. Differentiating the estimate with respect to [latex]ν[/latex] gives

[latex]ν(2 ν+3)-\left(ν^2+3 ν+2 ight)=0 \quad \Rightarrow \quad ν^2-2=0 \quad \Rightarrow \quad ν=\sqrt{2}[/latex]

Thus the least upper bound to be found with this parameterisation is

[latex]\omega^2 \leq \frac{c^2}{a^2} \frac{(\sqrt{2}+2)(\sqrt{2}+1)}{\sqrt{2}}=\frac{c^2}{2 a^2}(\sqrt{2}+2)^2 \quad \Rightarrow \quad \omega=(5.83)^{1 / 2} \frac{c}{a} .[/latex]

As noted, the actual lowest eigenfrequency is very little below this.

Question 22.21: A drumskin is stretched across a fixed circular rim of radiu...

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