8.3 Solutions
389
8.3.10 Waves in Pipes
8.83 L =
v
2 f
=
342
2 × 439
= 0.3895 m
The new frequency with the changed length L 1 is
f 1 = 439 + 2 = 441
f 2 = 439 − 2 = 437
L 1 =
v
2 f 1
=
342
2 × 441
= 0.3877 m
L 2 =
v
2 f 2
=
342
2 × 437
= 0.3913 m
The pipe must be shortened by 0.3895 − 0.3877 = 0.0018 m or 1.8 mm or
lengthened by 0.3913 − 0.3895 = 0.0018 m = 1.8 mm, so that 2 beats/s may
be heard when it is sounded with the fork.
8.84 Time taken for the plate to fall is
t =
2h
g
=
2 × 10
980
=
1
7
s
Time period
T =
t
8
=
1
7 × 8
=
1
56
s
Frequency, f =
1
T
= 56 Hz
8.85
v = v t = v 0
t + 273
273
= 33150
20 + 273
273
= 34343 cm/s
L = (2N + 1)
λ
4
=
(2N + 1)v
4 f 1
=
(2N + 1) × 34, 343
4 × 210
(N = 0, 1, 2) . . . (1)
Also L = (2M + 1)
v
4 f 2
=
(2M + 1) × 34343
4 × 350
(M = 0, 1, 2 . . .)
(2)
Equating right-hand sides of (1) and (2) and simplifying
(2N + 1)
(2M + 1)
=
3
5
(3)
The choice of N = 1 and M = 2 satisfies (3). Using N = 1 in (1) or M in (2)
gives L = 122.65 cm.