6.3 Solutions
273
6.3.3 Coupled Systems of Masses and Springs
6.40 Let spring 1 undergo an extension x 1 due to force F. Then x 1 =
F
k 1
. Similarly,
for spring 2, x 2 =
F
k 2
.
The force is the same in each spring, but the total displacement x is the sum
of individual displacements:
x = x 1 + x 2 =
F
k 1
+
F
k 2
k eq =
F
x
=
F
x 1 + x 2
=
F
F
k 1
+
F
k 2
=
1
1
k 1
+
1
k 2
=
k 1 k 2
k 1 + k 2
∴ T = 2π
m
k eq
= 2π
(k 1 + k 2 )m
k 1 k 2
6.41 The displacement is the same for both the springs and the total force is the
sum of individual forces.
F 1 = k 1 x, F 2 = k 2 x
F = F 1 + F 2 = (k 1 + k 2 )x
k eq =
F
x
= k 1 + k 2
T = 2π
m
k eq
= 2π
m
k 1 + k 2
6.42 Let the centre of mass be displaced by x. Then the net force
F = −k 1 x − k 2 x = −(k 1 + k 2 )x
Acceleration a =
F
m
= −(k 1 + k 2 )
x
m
= −ω 2 x
T =
2π
ω
= 2π
m
k 1 + k 2
6.43 Spring constant of the wire is given by
k
=
Y A
L
(1)