8.3 Solutions
385
Fig. 8.7
Maximum frequency will be heard when the whistle will be at A and minimum
when it is at B:
v s = ωr = 15 × 2 = 30 m/s
f max =
v f s
v − v s
=
330 × 540
330 − 30
= 594 Hz
f min =
v f s
v − v s
=
330 × 540
330 + 30
= 495 Hz
8.75 (a) The frequency of the rod is fixed, and so also for the air and the gas. The
distance between successive heaps of cork dust is equal to the distance
between two neighbouring nodes which is
1
2
λ, Fig. 8.8:
Fig. 8.8
v air = f λ air
v gas = f λ gas
∴ v gas = v air
λ gas
λ air
= 330 ×
10
8
= 425 m/s
(b) v rod = f λ rod =
v air
λ air
· λ rod =
340
2 × 0.08
(2L rod )
=
340
0.16
(2 × 1.2) = 5100 m/s
8.76 (i) v = f λ
∴ λ =
v
f
=
340
514
= 0.66 m
385
Fig. 8.7
Maximum frequency will be heard when the whistle will be at A and minimum
when it is at B:
v s = ωr = 15 × 2 = 30 m/s
f max =
v f s
v − v s
=
330 × 540
330 − 30
= 594 Hz
f min =
v f s
v − v s
=
330 × 540
330 + 30
= 495 Hz
8.75 (a) The frequency of the rod is fixed, and so also for the air and the gas. The
distance between successive heaps of cork dust is equal to the distance
between two neighbouring nodes which is
1
2
λ, Fig. 8.8:
Fig. 8.8
v air = f λ air
v gas = f λ gas
∴ v gas = v air
λ gas
λ air
= 330 ×
10
8
= 425 m/s
(b) v rod = f λ rod =
v air
λ air
· λ rod =
340
2 × 0.08
(2L rod )
=
340
0.16
(2 × 1.2) = 5100 m/s
8.76 (i) v = f λ
∴ λ =
v
f
=
340
514
= 0.66 m