266
6 Oscillations
(b) Let the two springs be stretched by equal force. Thus the work done
W A =
1
2
k A x
2
=
1
2
k A
F
k A
2
=
1
2
F 2
k A
W B =
1
2
F 2
k B
∴
W A
W B
=
k B
k A
Thus when two springs are stretched by the same force, less work will be
done on the stiffer spring.
Fig. 6.19
6.30 K trans + K rot + U = C = constant
1
2
mv
2
+
1
2
I ω
2
+ mg(R − r )(1 − cos θ) = C
Now I =
1
2
mr
2
ω =
v
r
3
4
m
dx
dt
2
+ mg(R − r )
θ 2
2
= C
Differentiating with respect to time
3
2
m
d 2 x
dt 2
dx
dt
+ mg(R − r )θ
dθ
dt
= 0
Now x = (R − r )θ
∴
3
2
d 2 x
dt 2 (R − r )
dθ
dt
+ gx
dθ
dt
= 0
Cancelling
dθ
dt
throughout