6.3 Solutions
257
Equation (1) becomes g = g 0
x
R
(3)
where x measures the distance from the centre. The acceleration g points opposite to the displacement x. We can therefore write
a = g = −
g 0 x
R
= −ω
2 x
(4)
with ω
2
=
g 0
R
Equation (4) shows that the box performs SHM. The period is calculated from
T =
2π
ω
= 2π
R
g 0
= 2π
6.4 × 10 6
9.8
= 5074 s or 84.6 min
6.7 Standard equation for SHM is
x = A sin(ωt + ε)
x = 4 sin
π t
3
+
π
6
(a) A = 4 cm
(b) ω =
π
3
. Therefore T =
2π
ω
= 6 s
(c) f =
1
T
=
1
6
/ s
(d) ε =
π
6
(e) v =
dx
dt
=
4π
3
cos
π t
3
+
π
6
=
4π
3
cos
π
3
× 1 +
π
6
= 0
(f) a =
dv
dt
= −
4π 2
9
sin
π
3
× 1 +
π
6
= −
4π 2
9
6.8 (a) K =
1
2
mω 2 (A 2 − x 2 ) U =
1
2
mω 2 x 2 K = U
∴
1
2
mω
2
(A
2
− x
2
) =
1
2
mω
2 x
2
∴ x =
A
√
2
(b) K =
1
2
mω 2
A 2 −
A 2
4
=
1
2
mω 2 3
4
A 2