Question 7.5: Three mass points are fixed equidistantly on a string that i...
Three mass points are fixed equidistantly on a string that is fixed at its endpoints.
(a) Determine the eigenfrequencies of this system if the string tension T can be considered constant (this holds for small amplitudes).
(b) Discuss the eigenvibrations of the system. Hint: Note Exercises 8.1 and 8.2 in Chap. 8.
(a) For the equations of motion of the system, one finds straightaway
m\ddot{x}_{1} + \left(\frac{2T}{L}\right) x_{1} − \left(\frac{T}{L}\right)x_{2} = 0,
m\ddot{x}_{2} +\left(\frac{2T}{L}\right) x_{2} −\left(\frac{T}{L}\right)x_{3} −\left(\frac{T}{L}\right)x_{1} = 0, (7.42)
m\ddot{x}_{3} +\left(\frac{2T}{L}\right) x_{3} −\left(\frac{T}{L}\right)x_{2} = 0.
Assuming periodic oscillations, i.e., solutions of the form
x_{1} = A sin(ωt +ψ), \quad \quad \ddot{x}_{1}=−ω^{2}A sin(ωt +ψ),x_{2} = B sin(ωt +ψ), \quad \quad \ddot{x}_{2}=−ω^{2}B sin(ωt +ψ),
x_{3} = C sin(ωt +ψ), \quad \quad \ddot{x}_{3}=−ω^{2}C sin(ωt +ψ),
we get after insertion into (7.42)
\left(\frac{2T}{L}−ω^{2}m\right) A − \left(\frac{T}{L}\right)B = 0,−\left(\frac{T}{L}\right)A +\left(\frac{2T}{L}−ω^{2}m\right)B − \left(\frac{T}{L}\right)C = 0,
−\left(\frac{T}{L} \right)B + \left(\frac{2T}{L}−ω^{2}m\right)C = 0. (7.43)
As in Exercise 8.2, one gets the equation for the frequencies of the system from the expansion of the determinant of coefficients:
\left(\frac{Lm}{T}\right)^{3}w^{6} −6 \left(\frac{Lm}{T}\right)^{2}ω^{4} + \frac{10Lm}{T}ω^{2} − 4 = 0or
\left(\frac{Lm}{T}\right)^{3}Ω^{3} −6 \left(\frac{Lm}{T}\right)^{2}Ω^{2} + \frac{10Lm}{T}Ω− 4 = 0 (7.44)
with Ω\widehat{=}ω^{2}. This cubic equation with the coefficients
a =\left(\frac{Lm}{T}\right)^{3}, \quad b=−6\left(\frac{Lm}{T}\right)^{2}, \quad c= \frac{10Lm}{T}, \quad d=−4can be solved by Cardano’s method.
With the substitutions
y =Ω + \frac{b}{3a}, \quad \quad 3p=−\frac{1}{3} \frac{b^{2}}{a^{2}}+ \frac{c}{a}=−2\frac{T^{2}}{L^{2}m^{2}} , \quad \quad 2q = \frac{2}{27} \frac{b^{3}}{a^{3}}− \frac{1}{3} \frac{bc}{a^{2}}+ \frac{d}{a}= 0we get q^{2} + p^{3} < 0, i.e., there are three real solutions which by using the auxiliary quantities
cos \varphi =− \frac{q}{\sqrt{-p^{3}}}= 0, \quad \quad y_{1} =−2\sqrt{−p} cos \left(\frac{\varphi}{3} − \frac{π}{3}\right) =− \sqrt{2} \frac{T}{Lm},
y_{2} =−2\sqrt{−p} cos \left(\frac{\varphi}{3} + \frac{π}{3}\right) = 0,
y_{3} = 2 \sqrt{−p} cos \frac{\varphi}{3}= \sqrt{2} \frac{T}{Lm}
can be calculated as
ω_{1} =\sqrt{0.6 \frac{T}{Lm}}, \quad \quad ω_{2} =\sqrt{ \frac{2T}{Lm}}, \quad \quad ω_{3} =\sqrt{3.4 \frac{T}{Lm}}.(b) From the first and third equation of (7.43), one finds for the amplitude ratios
\frac{B}{A}= \frac{B}{C} = 2− \frac{mLω^{2}}{T}. (7.45)
Discussion of the modes:
(1) ω = ω_{1} = (0.6T/Lm)^{1/2} inserted into (7.45) ⇒ B_{1}/A_{1} = B_{1}/C_{1} = 1.4 or B_{1} =1.4A_{1} = 1.4C_{1}.
All three masses are deflected in the same direction, where the first and third mass have equal amplitudes, and the second mass has a larger amplitude.
(2) ω = ω_{2} = (2T/Lm)^{1/2} inserted into (7.45) ⇒ B_{2}/A_{2} = B_{2}/C_{2} = 0 and A_{2} =−C_{2} from the second equation of (7.43). The central mass is at rest, while the first and third mass are vibrating in opposite directions but with equal amplitude.
(3) ω = ω_{3} = (3.4T/Lm)^{1/2} inserted into (7.45) ⇒ B_{3}/A_{3} = B_{3}/C_{3} = −1.4, i.e., A_{3} = C_{3} =−1.4B_{3}. The first and the last mass are deflected in the same direction, while the central mass vibrates with different amplitude in the opposite direction.
The system discussed here has three vibration modes with 0, 1, and 2 nodes, respectively. For a system with n mass points, both the number of modes as well as the number of possible nodes (n − 1) increases. A system with n→∞ is called a “vibrating string.”
A comparison of the figures clearly shows the approximation of the vibrating string by the system of three mass points.
