258
6 Oscillations
U =
1
2
mω
2 A 2
4
∴ K : U = 3 : 1
6.9 T = 2π
M
k
(1)
2 = 2π
M
k
(2)
3 = 2π
M + 2
k
(3)
Dividing (2) by (3) and solving for M, we get M = 1.6 kg.
6.10 a max = ω
2 A
5π
2
= ω
2 A
(1)
v = ω
A 2 − x 2
3π = ω
A 2 − 16
(2)
Solving (1) and (2), we get A = 5 cm and T =
2π
ω
=
2π
π
= 2 s.
6.11 α = ω
2 A
(1)
β = ω A
(2)
∴ β
2
= ω
2 A
2
= α A
or A =
β 2
α
Dividing (2) by (1)
β
α
=
1
ω
or T =
2π
ω
=
2πβ
α
6.12 By problem
mg + mv 2 /L
mg
= 1.01
∴
v 2
gL
= 0.01
Conservation of energy gives
1
2
mv 2 = mgh = mgL(1 − cos θ) mgL
θ 2
2
for small θ