Determine the eigenfrequencies of the system of three equal masses suspended between springs with the spring constant k, as is shown in Fig 8.7.
Hint: Consider the solution method of the preceding Exercise 8.2 and Mathematical Supplement 8.4.

Question

Accepted Answer

From Fig. 8.7, we extract for the equations of motion

[latex]m\ddot{x}_{1} =−kx_{1} −k(x_{1} −x_{2})−k(x_{1} −x_{3}),[/latex]

[latex]m\ddot{x}_{2} =−kx_{2} −k(x_{2} −x_{1})−k(x_{2} −x_{3}),[/latex]

[latex]m\ddot{x}_{3} =−kx_{3} −k(x_{3} −x_{1})−k(x_{3} −x_{2}),[/latex] (8.40)

or

[latex]m\ddot{x}_{1} +3kx_{1} − kx_{2} −kx_{3} = 0,[/latex]

[latex]m\ddot{x}_{2} +3kx_{2} − kx_{3} −kx_{1} = 0,[/latex]

[latex]m\ddot{x}_{3} +3kx_{3} − kx_{1} −kx_{2} = 0,[/latex] (8.41)

We look for the eigenvibrations. All mass points must vibrate with the same frequency. Thus, we adopt the ansatz

[latex]x_{1} = A cos(ωt +ψ), \ddot{x}_{1} =−ω^{2}A cos(ωt +ψ),[/latex]

[latex]x_{2} = B cos(ωt +ψ), \ddot{x}_{2} =−ω^{2}B cos(ωt +ψ),[/latex]

[latex]x_{3} = C cos(ωt +ψ), \ddot{x}_{3} =−ω^{2}C cos(ωt +ψ),[/latex]

and after insertion into (8.41), we get

[latex](3k −mω^{2})A −kB −kC = 0,[/latex]

[latex]−kA +(3k −mω^{2})B −kC = 0,[/latex]

[latex]−kA −kB +(3k −mω^{2})C = 0. [/latex] (8.42)

To get a nontrivial solution of (8.42), the determinant of coefficients must vanish:

[latex]\begin{vmatrix} (3k −mω^{2}) & -k & -k \ -k & (3k −mω^{2}) & -k \ -k & -k &(3k −mω^{2}) \end{vmatrix}=0 [/latex]

Expansion of the determinant leads to

[latex]0 = ω^{6} − \frac{9k}{m} ω^{4} + \frac{24k^{2}}{m^{2}}ω^{2} − \frac{16k^{3}}{m^{3}}[/latex]

or

[latex]0 = Ω^{3} − \frac{9k}{m} Ω^{2} + \frac{24k^{2}}{m^{2}}Ω − \frac{16k^{3}}{m^{3}}[/latex]

where we substituted [latex] Ω= ω^{2}[/latex] (see Exercise 8.2). The general cubic equation [latex]aΩ^{3} +bΩ^{2} + cΩ + d = 0[/latex] (in our case [latex]a = 1, b =−9k/m, c = 24k^{2}/m^{2}, d =−16k^{3}/m^{3})[/latex] can according to Mathematical Supplement 8.4 be reduced to

[latex]y^{3} +3py +2q = 0,[/latex]

where

[latex]y = Ω + \frac{b}{3a}, 3p=−\frac{1}{3} \frac{b^{2}}{a^{2}} + \frac{c}{a}, 2q = \frac{2}{27} \frac{b^{3}}{a^{3}}− \frac{1}{3} \frac{bc}{a^{2}} + \frac{d}{a}.[/latex]

Insertion leads to

[latex]3p=−3 \frac{k^{2}}{m^{2}} , 2q = 2 \frac{k^{3}}{m^{3}} ⇒ q^{2} +p^{3} = 0,[/latex]

i.e., there exist 3 solutions (the real roots); 2 of them coincide. Hence, the vibrating system being treated here is degenerate. As in Exercise 8.2, the solutions can be calculated using tabulated auxiliary quantities. For these, we obtain

[latex]cos \varphi =\frac{−q}{\sqrt{−p^{3}}}, y_{1} = 2 \sqrt{−p} cos \frac{\varphi}{3},[/latex]

[latex]y_{2} =−2 \sqrt{−p} cos \left(\frac{\varphi}{3} + \frac{π}{3} ight),[/latex]

[latex]y_{3} =−2 \sqrt{−p} cos \left(\frac{\varphi}{3} - \frac{π}{3} ight),[/latex]

and, after insertion, for the eigenfrequencies of the system

[latex]ω_{3} =\sqrt{\frac{k}{m}}, ω_{1} = ω_{2} = 2 \sqrt{\frac{k}{m}}.[/latex]

Question 8.3: Determine the eigenfrequencies of the system of three equal ...

Related Answered Questions

Calculate the eigenfrequencies of the system of three different masses that are fixed equidistantly on a stretched string, as is shown in Fig. 8.5. Hint: For small amplitudes, the string tension T does not change! ...

Verified Answer:

Consider a string of density σ that is stretched between two points and is excited with small amplitudes. (a) Calculate in general the kinetic and potential energy of the string. (b) Calculate the kinetic and potential energy for waves of the form ...

Verified Answer: