6.3 Solutions
265
(b) The time period
T =
2π
ω
= 2π
ml
AP
6.27 y = 8 sin
2π t
T
+ φ
At t = 0; 4 = 8 sin φ
∴ φ = 30
◦
=
π
6
y = 8 sin
2π × 6
24
+
π
6
= 8 sin 120 = 4
√
3 cm
6.28 Time period of a loaded spring
T = 2π
M +
m
3
k
(1)
where M is the suspended mass, m is the mass of the spring and k is the spring
constant
0.89 = 2π
1.5 +
m
3
k
(2)
1.13 = 2π
2.5 +
m
3
k
(3)
Dividing the two equations and solving for m, we get m = 0.39 kg.
6.29 (a) k A > k B
Let the springs be stretched by the same amount. Then the work done on
the two springs will be
W A =
1
2
k A x
2 W B =
1
2
k B x
2
W A
W B
=
k A
k B
Thus W A > W B , i.e. when two springs are stretched by the same amount,
more work will be done on the stiffer spring.