Classical Dynamics of Particles and Systems
Classical Dynamics of Particles and Systems
5th Edition
ISBN: 9780534408961
Author: Stephen T. Thornton, Jerry B. Marion
Publisher: Cengage Learning
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Chapter 12, Problem 12.1P
To determine

The characteristic frequencies of the two coupled oscillators with all spring constants different, and to compare it with the natural frequencies of the two oscillators in the absence of coupling.

Expert Solution & Answer
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Answer to Problem 12.1P

The characteristic frequencies of the two coupled oscillators with all spring constants different are ω+=(k1+k2+2k122M+12M(k1k2)2+4k122)1/2 and ω=(k1+k2+2k122M12M(k1k2)2+4k122)1/2. In the absence of coupling ω+>ω01 and ω<ω02 where ω01 and ω02 are the frequencies when mass m2 and m1 are held fixed respectively.

Explanation of Solution

The system of the two coupled oscillators with all spring constants different are shown in Figure 1.

Classical Dynamics of Particles and Systems, Chapter 12, Problem 12.1P

In the figure, the spring constants of the springs are marked as k1, k2, and k12.

Write the equations of motions for the system.

  Mx¨1+(k1+k12)x1k12x2=0Mx¨2+(k2+k12)x2k12x1=0}        (I)

Let us attempt the solutions of the form;

  x1(t)=B1eiωtx2(t)=B2eiωt}        (II)

Here, B1 and B2 are constants.

Use the solutions in (II) in (I).

  (k1+k12Mω2)B1k12B2=0k12B1+(k2+k12Mω2)B2=0}        (III)

The condition for obtaining a non-trivial solution is that, the determinant of the coefficients of B1 and B2 must be zero. Thus, it yields;

  [k1+(k12Mω2)][k2+(k12Mω2)]=k122        (IV)

Solve equation (IV) to get an expression for ω2.

  ω2=k1+k2+2k122M±12M(k1k2)2+4k122        (V)

Thus, the characteristic frequencies will be;

  ω+=(k1+k2+2k122M+12M(k1k2)2+4k122)1/2

  ω=(k1+k2+2k122M12M(k1k2)2+4k122)1/2

Consider the case; k1=k2=k, then equation (V) reduces to;

  ω2=(k+k12±k12)M

If the mass m2 is not moving, then the frequency of oscillation of m1 will be;

  ω012=1M(k1+k12)        (VI)

If the mass m1 is not moving, then the frequency of oscillation of m2 will be;

  ω022=1M(k2+k12)        (VII)

Comparing equation (VI) and (VII) with the two frequencies given by equation (V), it can be observed that;

  ω+2=12M[k1+k2+2k12+(k1k2)2+4k122]ω+2>12M[k1+k2+2k12+(k1k2)]ω+2>12M[2k1+2k12]ω+2>1M[k1+k12]ω+2>ω012        (VIII)

Similarly;

  ω2=12M[k1+k2+2k12(k1k2)2+4k122]ω+2<12M[k1+k2+2k12(k1k2)]ω+2<12M[2k2+2k12]ω+2<1M[k2+k12]ω+2<ω022        (IX)

From equation (IX) and (X) it is clear that;

  ω+>ω01ω<ω02

If k1>k2, then the ordering of the frequencies will be as;

  ω+>ω01>ω02>ω

Conclusion:

Therefore, the characteristic frequencies of the two coupled oscillators with all spring constants different are ω+=(k1+k2+2k122M+12M(k1k2)2+4k122)1/2 and ω=(k1+k2+2k122M12M(k1k2)2+4k122)1/2. In the absence of coupling ω+>ω01 and ω<ω02 where ω01 and ω02 are the frequencies when mass m2 and m1 are held fixed respectively.

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