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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!

8.5
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From Fig. 8.6, we extract for the equations of motion

2mx¨1=T[(x2−x1)L]−T[x1L],2m\ddot{x}_{1} = T \left[ \frac{(x_{2} − x_{1})}{L}\right]−T \left[\frac{x_{1}}{L}\right],

mx¨2=−T[(x2−x1)L]−T[(x2−x3)L],m\ddot{x}_{2} =−T \left[ \frac{(x_{2} − x_{1})}{L}\right] − T \left[ \frac{(x_{2} − x_{3})}{L}\right] ,                    (8.36)

3mx¨3 =T[(x2−x3)L]−T [x3L],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

x1=Asin(ωt+ψ),                       x¨1=−ω2A sin(ωt+ψ),x_{1} = A sin(ωt +ψ),                        \ddot{x}_{1} =−ω^{2}A  sin(ωt + ψ),

 

x2=Bsin(ωt+ψ),                       x¨2=−ω2B sin(ωt+ψ),x_{2} = B sin(ωt +ψ),                        \ddot{x}_{2} =−ω^{2}B  sin(ωt + ψ),

 

x3=Csin(ωt+ψ),                       x¨3=−ω2C 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

(2TL−2mω2)A−(TL)B=0,\left(\frac{2T}{L}− 2mω^{2}\right) A − \left(\frac{T}{L}\right)B = 0,

−(TL)A+(2TL−mω2)B−(TL)C=0,− \left(\frac{T}{L}\right)A + \left(\frac{2T}{L}− mω^{2}\right)B − \left(\frac{T}{L}\right)C = 0,                           (8.37)

−(TL)B+(2TL−3mω2)C=0.− \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:

∣(2TL−2mω2)−TL0−TL(2TL−mω2)−TL0−TL (2TL−3mω2)∣=0\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}=0

Expansion of the determinant leads to

0=6m3ω6−(22Tm2L)ω4+(19T2mL2)ω2−(4T3L3)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=6m3Ω3−(22Tm2L)Ω2+(19T2mL2)Ω−(4T3L3)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=ω2Ω= ω^{2}. This leads to the cubic equation

aΩ3+bΩ2+cΩ+d=0,aΩ^{3} +bΩ^{2} + cΩ+d = 0,

where

a=6m3,      b=−22Tm2L,       c=19T2mL2,             d=−4T3L3.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)

y3+3py+2q=0,y^{3} +3py +2q = 0,                                (8.39)

where

y=Ω+b3a=Ω−119TLmy =Ω + \frac{b}{3a} = Ω − \frac{11}{9} \frac{T}{Lm}

and

3p=−13b2a2+ca  and      2q=227b3a3−13bca2+da.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=−7154T2L2m2,          2q=−6531458T3L3m3.3p=−\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

q2+p3<0,q^{2} +p^{3} < 0,

i.e., there exist 3 real solutions of the cubic equation (8.39).

For the case q2+p3≤0q^{2} +p^{3} ≤ 0, the solutions y1,y2,y3y_{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 φ=−q−p3,        y1=2−p cosφ3,cos  \varphi =\frac{−q}{\sqrt{−p^{3}}},                y_{1} = 2\sqrt{−p}  cos \frac{\varphi}{3},

 

y2=−2−p cos(φ3+π3),y_{2} =−2 \sqrt{−p}  cos \left(\frac{\varphi}{3}+ \frac{π}{3}\right),

 

y3=−2−p cos(φ3−π3),y_{3} =−2 \sqrt{−p}  cos \left(\frac{\varphi}{3}- \frac{π}{3}\right),

and finally, for the eigenfrequencies of the system

ω1=0.563TLm,     ω2=0.916TLm,           ω3=1.585 TLm.ω_{1} = 0.563 \sqrt{\frac{T}{Lm}},           ω_{2} = 0.916 \sqrt{\frac{T}{Lm}} ,                       ω_{3} = 1.585  \sqrt{\frac{T}{Lm}}.
8.6

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