When the masses of the coupled pendulum of Figure 11.1 are no longer equal the e

Question

When the masses of the coupled pendulum of Figure 11.1 are no longer equal the equations of motion become m1 = -m1( b)x - k(x - y) m2 = -m2 ( b)y + k(x - y) Show that we can choose the normal coordinates X = m1x + m2y/m1 + m2 with a normal mode frequency omega12 = g/b and Y = x - y with a normal mode frequency omega22 = g/b + k (1/m1 + 1/m2). Note that X is the coordinate of the center of mass of the system while the effective mass in the Y mode is the reduced mass mu of the system (1/mu = 1/m1 + 1/m2).

Explanation / Answer

X = (m1 x + m2 y)/(m1 + m2) <-------------------equation1

>>>> X'' = (m1 x'' + m2 y'')/(m1 + m2) <----------equation2

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Y = x - y <------------------------------equation3

>>>> Y'' = x'' - y''   <-------------------equation4

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m1 x'' = -m1 (g/b) x - k (x-y)

m2 y'' = -m2 (g/b) y + k (x-y)

>>>> add the two equations:

m1 x'' + m2 y'' = -m1 (g/b) x - m2 (g/b) y

m1 x'' + m2 y'' = -(g/b)(m1 x + m2 y)

(m1 x'' + m2 y'')/(m1 + m2) = -(g/b)(m1 x + m2 y)/(m1 + m2)

>>>> from equations 1 and 2:

X'' = -(g/b) X

>>>> 2 = g/b

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m1 x'' = -m1 (g/b) x - k (x-y) >>> x'' = -(g/b) x - k/m1 (x-y)

m2 y'' = -m2 (g/b) y + k (x-y) >>> y'' = -(g/b) y + k/m2 (x-y)

>>>> substitude the two equations:

x'' - y'' = -(g/b) (x - y) - k (x - y) (1/m1 + 1/m2)

>>>> from equations 3 and 4:

Y'' = (-(g/b)-k(1/m1 + 1/m2)) Y

Y'' = -((g/b)+k(1/m1 + 1/m2)) Y

>>> 2 = (g/b)+k(1/m1 + 1/m2)

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