262
6 Oscillations
Cancelling dx/dt throughout and simplifying
d 2 x
dt 2 +
2k
3m
x = 0
This is the equation for SHM
with ω
2
=
2k
3m
T =
2π
ω
= 2π
3m
2k
6.22 The time period of the pendulums is
T 1 = 2π
60
g
(1)
T 2 = 2π
63
g
(2)
Let the time be t in which the longer length pendulum makes n oscillations
while the shorter one makes (n + 1) oscillations. Then
t = (n + 1)T 1 = nT 2
(3)
Using (1) and (2) in (3), we find n = 40.5 and t = 64.49 s.
6.23 Let g 0 be the acceleration due to gravity on the ground and g at height above
the ground. Then
g =
g 0 R 2
(R + h) 2
At the ground, T 0 = 2π
L
g 0
. At height h, T = 2π
L
g
T = T 0
g 0
g
= T 0
1 +
h
R
= 2
1 +
320
6.4 × 10 6
= 2.0001 s
Time lost in one oscillation on the top of the tower = 2.0001 − 2.0000 =
0.0001 s. Number of oscillations in a day for the pendulum which beats
seconds on the ground
=
86400
2.0
= 43,200