Question 7.1: Two equal masses move without friction on a plate. They are ...
Two equal masses move without friction on a plate. They are connected to each other and to the wall by two springs, as is indicated by Fig. 7.3. The two spring constants are equal, and the motion shall be restricted to a straight line (onedimensional motion). Two equal masses coupled by two equal springs.
Find
(a) the equations of motion,
(b) the normal frequencies, and
(c) the amplitude ratios of the normal vibrations and the general solution.
(a) Let x_{1} and x_{2} be the displacements from the rest positions. The equations of motion then read
m\ddot{x}_{1} =−kx_{1} +k(x_{2} −x_{1}), (7.9)
m\ddot{x}_{2} =−k(x_{2} − x_{1}). (7.10)
(b) For determining the normal frequencies, we use the ansatz
x_{1} = A_{1} cos ωt, x_{2} = A_{2} cos ωtand thereby get from (7.9) and (7.10) the equations
(2k −mω^{2})A_{1} − kA_{2} = 0,−kA_{1} +(k −mω^{2})A_{2} = 0. (7.11)
From the requirement for nontrivial solutions of the system of equations, it follows that the determinant of coefficients vanishes:
D =\begin{vmatrix} 2k − mω^{2} & −k \\ −k & k− mω^{2} \end{vmatrix}=0From this follows the determining equation for the eigenfrequencies,
ω^{4} − 3 \frac{k}{m} ω^{2} + \frac{k^{2}}{m^{2}}= 0,with the positive solutions
ω_{1} =\frac{\sqrt{5}+1}{2} \sqrt{\frac{k}{m}} and ω_{2} =\frac{\sqrt{5}-1}{2} \sqrt{\frac{k}{m}} , ω_{1} >ω_{2}.(c) By inserting the eigenfrequencies in (7.11) one sees that the higher frequency ω_{1} corresponds to the opposite-phase mode, and the lower frequency ω_{2} to the equalphase normal vibration:
with ω^{2}_{1} = \frac{1}{2} (3 +\sqrt{5}) \frac{k}{m}, it follows from (7.11) that A_{2} =− \frac{\sqrt{5}−1}{2}{A_{1}},
with ω^{2}_{2} = \frac{1}{2} (3 − \sqrt{5}) \frac{k}{m}, it follows from (7.11) that A_{2} =\frac{\sqrt{5}+ 1}{2} A_{1}.
Since the two mass points are fixed in different ways, we find amplitudes of different magnitudes.
The general solution is obtained as a superposition of the normal vibrations, using the calculated amplitude ratios:
x_{1}(t) = C_{1} cos(ω_{1}t +\varphi_{1})+C_{2} cos(ω_{2}t +\varphi_{2}),x_{2}(t)=− \frac{\sqrt{5}− 1}{2} C_{1} cos(ω_{1}t +\varphi_{1})+ \frac{\sqrt{5} +1}{2} C_{2} cos(ω_{2}t +\varphi_{2}).
The 4 free constants are determined from the initial conditions of the specific case.