Duke Vuong MEAE 7010 Homework # 4

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Duke Vuong
MEAE 7010
Homework # 4
5. Suppose that the uniform string considered in Section 5.2 is unrestrained from
transverse motion at the end x = 0, and is attached at the end x = l to a yielding
support of modulus k, the ends o the string being constrained against appreciable
movement parallel to the axis of rotation. Show that the nth critical speed is given by
n 
n
T
l

where cot  n  n and  
T
.
kl
Solution:
From the differential equation (9) of the form,
d2y
 y  0
dx 2
with the general solution that satisfies the end condition y (0)  0 (at x = 0)
y  C sin  x .
The second end (x = l) is attached to a yielding support of modulus k
y (l )  ly (l )
with
y ( x)  C  cos  x .
We have
sin  l    l cos  l .
With the dimensionless parameter  , of the form
  l
The characteristic values of  are given by
n 
 n2
l2
Per equation (8), the constant parameter is

 2
T
.
Thus, the corresponding critical speeds are given by
n 
n
T
l

.
22. Reduce each of the following differential equations to the standard form
d  dy 
 p   q  r  y  0
dx  dx 
(a) x
d2y
dy
 2  x    y  0
2
dx
dx
Solution:
Setting
a
pe
 a10 dx
q
a2
p
a0
r
a3
p.
a0
With
a0  x
a1  2
a2  x
a3  1 .
We have
p  x2
r  x.
q  x2
Therefore, the reduced equation has the form


d  2 dy 
2
x
  x  x y  0
dx  dx 
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