6.3 Solutions
255
6.2 (a) v = ω
A 2 − x 2
(1)
16 = ω
A 2 − 3 2
(2)
12 = ω
A 2 − 4 2
(3)
Solving (2) and (3) A = 5 cm and ω = 4 rad/s
(b) Therefore T =
2π
ω
=
2π
4
= 1.57 s
6.3 x = A sin ωt
v =
dx
dt
= ω A cos ωt
v max = Aω =
2π A
T
=
2π × 5
2
= 5π cm/s
At the equilibrium position the weight of the bob and the tension act in the
same direction
Tension = mg +
mv 2
max
L
Now the length of the simple pendulum is calculated from its period T .
L =
gT 2
4π 2 =
980 × 2 2
4π 2 = 99.29 cm
Tension = m
1 +
v 2
max
gL
g = 50
1 +
25π 2
980 × 99.29
g
= 50.13 g dynes = 50.13 g wt
6.4 The general equation of SHM is
x = A sin(ωt + ε)
ω =
2π
T
=
2π
16
=
π
8
When t = 2 s, x = 0.
0 = A sin
π
8
× 2 + ε
Since A = 0, sin
π
4
+ ε
= 0