8.3 Solutions
371
The total energy of the entire string is obtained by integrating from 0 to L
E n =
dE n =
1
2
μω
2
n (a
2
n + b
2
n )
L
0
sin
2
(k n x)dx
Now
L
0
sin
2
(k n x)dx =
L
0
sin
2
nπ x
L
dx =
L
2
∴ E n =
1
4
μLω
2
n (a
2
n + b
2
n ) =
1
4
Mω
2
n (a
2
n + b
2
n )
where M is the total mass of the string.
(b) For the string plucked at the centre a n =
8h
n 2 π 2 (see prob. 8.3). Further
ω n = k n v =
nπv
L
and b n = 0. Thus the energy of vibration
E n =
M
4
nπv
L
2
8h
n 2 π 2
2
=
16Mh 2 v 2
n 2 π 2 L 2
∴
E 1
E 3
=
9
1
8.3.2 Waves in Solids
8.32 (a) For the rod clamped at one end and free at the other (fixed–free)
f n =
n
4L
γ
ρ
(n = 1, 3, 5, . . .)
f 1 =
1
4L
γ
ρ
=
1
4 × 0.25
2 × 10 11
7860
= 5044 Hz
(b) (i) For the rod free at both ends (free–free)
f n =
n
2L
Y
ρ
(n = 1, 2, 3, . . .)
(ii) For the rod clamped at the midpoint
f n =
n
2L
Y
ρ
(n = 1, 3, 5, . . .)
(iii) For the bar clamped at both ends (fixed–fixed)
f n =
n
2L
Y
ρ
(n = 1, 2, 3, . . .)