Coupled Harmonic Oscillators X, =0 x= x' = 2t Derivative X= x" = 2nd Derivative ki k2 k3 we M A) Write down the 2nd law for each of the masses. Use coordinates x, and x2 for m and M, and % X, and a,92.

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please show all work for parts A-C. :)

Coupled Harmonic Oscillators
X, 30
k3
M
*= x' = 2t Derivative
X= x" = 2nd Derivative
ki
wee
A) Write down the 2nd law for each of the masses.
Use coordinates x. and xz for m and M, and *, X2 and a,,a2.
Hìnt : the force from spring I on m only depends on X,, but the
force from spring 2 on m depends on (x2-x1).
B) To simplify, let k, = k3 =k and m= M. Rownite
cquations from parrt A) right abore each other.
C) Define two new variabks , X( Greek "chi") =x. +xa and AX =x-X2.
Then add and subtract your two differential evahions to produce two new,
but very simple (harmonic) ones.
D) Write down the solution to both egvations using wo=/m, wa=ktak.
and coefficients A, B,C, and D.
E) Now assume k=l0k2 (this means that the two masses are "weakly coupled").
Also assume x,(0) = -10cm, i,l0) =0, X. (0) =0, x1 (0) = 0. Solve for
A, B,C,D, and solve for x,lt).
F) Look up a trig. identity to show that your solution from part E) can
be
two differential
your
written in the form :
x,(t) =
= -10cm cos ( Wotwat Cas W2-
2
G) For k= loN/m, m= 1.00 kg graph the behavior of x,(t) from
t=0 to t= 1T
W2-Wo
H) Based on your answer to part W guess what x(t) looks like.
Try to guess X2(t)'s functional form.
Transcribed Image Text:Coupled Harmonic Oscillators X, 30 k3 M *= x' = 2t Derivative X= x" = 2nd Derivative ki wee A) Write down the 2nd law for each of the masses. Use coordinates x. and xz for m and M, and *, X2 and a,,a2. Hìnt : the force from spring I on m only depends on X,, but the force from spring 2 on m depends on (x2-x1). B) To simplify, let k, = k3 =k and m= M. Rownite cquations from parrt A) right abore each other. C) Define two new variabks , X( Greek "chi") =x. +xa and AX =x-X2. Then add and subtract your two differential evahions to produce two new, but very simple (harmonic) ones. D) Write down the solution to both egvations using wo=/m, wa=ktak. and coefficients A, B,C, and D. E) Now assume k=l0k2 (this means that the two masses are "weakly coupled"). Also assume x,(0) = -10cm, i,l0) =0, X. (0) =0, x1 (0) = 0. Solve for A, B,C,D, and solve for x,lt). F) Look up a trig. identity to show that your solution from part E) can be two differential your written in the form : x,(t) = = -10cm cos ( Wotwat Cas W2- 2 G) For k= loN/m, m= 1.00 kg graph the behavior of x,(t) from t=0 to t= 1T W2-Wo H) Based on your answer to part W guess what x(t) looks like. Try to guess X2(t)'s functional form.
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