2.3 Solutions
99
Calling β 2 = g/L, (3) becomes
d 2 y
dt 2 − β
2 y = 0
( 4 )
which has the solution
y = Ce
βt
+ De
−βt
(5)
where C and D are constants. When t = 0, y = y 0
∴ y 0 = C + D
(6)
Further
dy
dt
= β
Ce
βt
− De
−βt
When t = 0,
dy
dt
= 0
∴ 0 = C − D
∴ C = D =
y 0
2
(7)
Using (7) in (5)
y =
y 0
2
e
βt
+ e
−βt
(8)
Thus the complete solution is
y = y 0 cosh(βt)
(9)
Note that initially both the terms in the parenthesis of (8) are important. As t
increases, the second term becomes vanishingly small and the first term alone
dominates. Thus y (length of the rope hanging down) increases exponentially
with time.
From (3) the acceleration
d 2 y
dt 2 =
g
L
y
Thus acceleration continuously increases with increasing value of y. This then
is the case of non-uniform acceleration.
99
Calling β 2 = g/L, (3) becomes
d 2 y
dt 2 − β
2 y = 0
( 4 )
which has the solution
y = Ce
βt
+ De
−βt
(5)
where C and D are constants. When t = 0, y = y 0
∴ y 0 = C + D
(6)
Further
dy
dt
= β
Ce
βt
− De
−βt
When t = 0,
dy
dt
= 0
∴ 0 = C − D
∴ C = D =
y 0
2
(7)
Using (7) in (5)
y =
y 0
2
e
βt
+ e
−βt
(8)
Thus the complete solution is
y = y 0 cosh(βt)
(9)
Note that initially both the terms in the parenthesis of (8) are important. As t
increases, the second term becomes vanishingly small and the first term alone
dominates. Thus y (length of the rope hanging down) increases exponentially
with time.
From (3) the acceleration
d 2 y
dt 2 =
g
L
y
Thus acceleration continuously increases with increasing value of y. This then
is the case of non-uniform acceleration.