Question 8.2: Calculate the eigenfrequencies of the system of three differ...
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!
From Fig. 8.6, we extract for the equations of motion
2m\ddot{x}_{1} = T \left[ \frac{(x_{2} − x_{1})}{L}\right]−T \left[\frac{x_{1}}{L}\right],m\ddot{x}_{2} =−T \left[ \frac{(x_{2} − x_{1})}{L}\right] − T \left[ \frac{(x_{2} − x_{3})}{L}\right] , (8.36)
3m\ddot{x}_{3} = T \left[ \frac{(x_{2} − x_{3})}{L}\right] −T \left[\frac{x_{3}}{L}\right],We look for the eigenvibrations. All mass points must then vibrate with the same frequency. We therefore start with
x_{1} = A sin(ωt +ψ), \ddot{x}_{1} =−ω^{2}A sin(ωt + ψ),x_{2} = B sin(ωt +ψ), \ddot{x}_{2} =−ω^{2}B sin(ωt + ψ),
x_{3} = C sin(ωt +ψ), \ddot{x}_{3} =−ω^{2}C sin(ωt + ψ).
Hence, after insertion into equation (8.36) one gets
\left(\frac{2T}{L}− 2mω^{2}\right) A − \left(\frac{T}{L}\right)B = 0,− \left(\frac{T}{L}\right)A + \left(\frac{2T}{L}− mω^{2}\right)B − \left(\frac{T}{L}\right)C = 0, (8.37)
− \left(\frac{T}{L}\right)B + \left(\frac{2T}{L}− 3mω^{2}\right) C = 0.For evaluating the eigenfrequencies of the system, i.e., for solving equation (8.37), the determinant of coefficients must vanish:
\begin{vmatrix} \left(2\frac{T}{L}− 2mω^{2}\right) & -\frac{T}{L} & 0 \\ -\frac{T}{L} & \left(2\frac{T}{L}− mω^{2}\right) & -\frac{T}{L} \\ 0 & -\frac{T}{L} & \left(2\frac{T}{L}− 3mω^{2}\right)\end{vmatrix}=0Expansion of the determinant leads to
0 = 6m^{3}ω^{6} − \left( \frac{22Tm^{2}}{L} \right) ω^{4} + \left( \frac{19T^{2}m}{L^{2}}\right)ω^{2} − \left(\frac{4T^{3}}{L^{3}}\right)or
0 = 6m^{3}Ω^{3} − \left( \frac{22Tm^{2}}{L} \right) Ω^{2} + \left( \frac{19T^{2}m}{L^{2}}\right)Ω − \left(\frac{4T^{3}}{L^{3}}\right) (8.38)
where we substituted Ω= ω^{2}. This leads to the cubic equation
aΩ^{3} +bΩ^{2} + cΩ+d = 0,where
a = 6m^{3}, b=\frac{−22Tm^{2}}{L}, c=\frac{ 19T^{2}m}{L^{2}} , d=\frac{−4T^{3}}{L^{3}} .It can be transformed to the representation (reduction of the cubic equation)
y^{3} +3py +2q = 0, (8.39)
where
y =Ω + \frac{b}{3a} = Ω − \frac{11}{9} \frac{T}{Lm}and
3p=−\frac{1}{3} \frac{b^{2}}{a^{2}} + \frac{c}{a} and 2q = \frac{2}{27} \frac{b^{3}}{a^{3}}− \frac{1}{3} \frac{bc}{a^{2}}+ \frac{d}{a}.Insertion leads to
3p=−\frac{71}{54} \frac{T^{2}}{L^{2}m^{2}} , 2q =− \frac{653}{1458} \frac{T^{3}}{L^{3}m^{3}} .From this, it follows that
q^{2} +p^{3} < 0,i.e., there exist 3 real solutions of the cubic equation (8.39).
For the case q^{2} +p^{3} ≤ 0, the solutions y_{1}, y_{2}, y_{3} can be calculated using tabulated auxiliary quantities (see Mathematical Supplement 8.4). Direct application of Cardano’s formula would lead to complex expressions for the real roots, hence the above method is convenient.
After insertion one obtains for the auxiliary quantities
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 finally, for the eigenfrequencies of the system
ω_{1} = 0.563 \sqrt{\frac{T}{Lm}}, ω_{2} = 0.916 \sqrt{\frac{T}{Lm}} , ω_{3} = 1.585 \sqrt{\frac{T}{Lm}}.