Section 3.9– Number 21
Writing g for the acceleration due to gravity, the period, T, of a pendulum of
length l is given by:
l
g
T = 2π
a)
Show that if the length of the pendulum changes by ∆l , the change in the
period, ∆T , is given by
∆T =
T
∆l
2l
Solution:
First, we need to rewrite this formula "separating" the constants from the
variables.
1
l2
T = 2π
=
g
dT
dl
Finding
2π
g
1
l2
, we get:
dT
2π  1 − 21 
π 1
=
i
 l =
dl
2
g
g l

Or
dT =
π
1
g
i
l
dl
This is equivalent to
∆T =
π
g
1
i
l
∆l
Recall that T =
1
T g = 2π l 2
2π
g
1
l2 .
Multiplying both sides by g , we get:
Solving for g , we get:
1
2π l 2
g =
T
Substituting this into the ∆T formula, we get:
∆T =
π
2π l
T
1
1
2
i
Thus ∆T =
l
1
2
∆l = π i
T
2π l
1
1
2
i
1
∆l
l2
T
∆l .
2l
b)
If the length of the pendulum increases by 2%, by what percent does the
period change?
Solution:
We have that ∆l = 0.02. Substitute this into the ∆T formula. So
∆T =
T
T
(0.02) = (0.01)
2l
l
Thus the period change is 1%.