Question 8.3: Determine the eigenfrequencies of the system of three equal ...
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.
From Fig. 8.7, we extract for the equations of motion
m\ddot{x}_{1} =−kx_{1} −k(x_{1} −x_{2})−k(x_{1} −x_{3}),m\ddot{x}_{2} =−kx_{2} −k(x_{2} −x_{1})−k(x_{2} −x_{3}),
m\ddot{x}_{3} =−kx_{3} −k(x_{3} −x_{1})−k(x_{3} −x_{2}), (8.40)
or
m\ddot{x}_{1} +3kx_{1} − kx_{2} −kx_{3} = 0,m\ddot{x}_{2} +3kx_{2} − kx_{3} −kx_{1} = 0,
m\ddot{x}_{3} +3kx_{3} − kx_{1} −kx_{2} = 0, (8.41)
We look for the eigenvibrations. All mass points must vibrate with the same frequency. Thus, we adopt the ansatz
x_{1} = A cos(ωt +ψ), \ddot{x}_{1} =−ω^{2}A cos(ωt +ψ),x_{2} = B cos(ωt +ψ), \ddot{x}_{2} =−ω^{2}B cos(ωt +ψ),
x_{3} = C cos(ωt +ψ), \ddot{x}_{3} =−ω^{2}C cos(ωt +ψ),
and after insertion into (8.41), we get
(3k −mω^{2})A −kB −kC = 0,−kA +(3k −mω^{2})B −kC = 0,
−kA −kB +(3k −mω^{2})C = 0. (8.42)
To get a nontrivial solution of (8.42), the determinant of coefficients must vanish:
\begin{vmatrix} (3k −mω^{2}) & -k & -k \\ -k & (3k −mω^{2}) & -k \\ -k & -k &(3k −mω^{2}) \end{vmatrix}=0Expansion of the determinant leads to
0 = ω^{6} − \frac{9k}{m} ω^{4} + \frac{24k^{2}}{m^{2}}ω^{2} − \frac{16k^{3}}{m^{3}}or
0 = Ω^{3} − \frac{9k}{m} Ω^{2} + \frac{24k^{2}}{m^{2}}Ω − \frac{16k^{3}}{m^{3}}where we substituted Ω= ω^{2} (see Exercise 8.2). The general cubic equation aΩ^{3} +bΩ^{2} + cΩ + d = 0 (in our case a = 1, b =−9k/m, c = 24k^{2}/m^{2}, d =−16k^{3}/m^{3}) can according to Mathematical Supplement 8.4 be reduced to
y^{3} +3py +2q = 0,where
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}.Insertion leads to
3p=−3 \frac{k^{2}}{m^{2}} , 2q = 2 \frac{k^{3}}{m^{3}} ⇒ q^{2} +p^{3} = 0,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
cos \varphi =\frac{−q}{\sqrt{−p^{3}}}, y_{1} = 2 \sqrt{−p} cos \frac{\varphi}{3},y_{2} =−2 \sqrt{−p} cos \left(\frac{\varphi}{3} + \frac{π}{3}\right),
y_{3} =−2 \sqrt{−p} cos \left(\frac{\varphi}{3} – \frac{π}{3}\right),
and, after insertion, for the eigenfrequencies of the system
ω_{3} =\sqrt{\frac{k}{m}}, ω_{1} = ω_{2} = 2 \sqrt{\frac{k}{m}}.