388
8 Waves
8.3.9 Beat Frequency
8.81 When the man moves towards the source
f 0 =
(v + v 0 ) f s
v
When the man moves away from the source
f
0 =
(v − v 0 ) f
s
v
∴ f 0 − f
0 =
v 0
v
( f s + f
s ) + f s − f
s
=
1.5
330
(548 + 552) + 548 − 552 = 1 s
∴ Beat frequency = 1/s.
8.82 Suppose the frequency of the unknown fork (unloaded) is n. Then n = 300±2
Case (i) Suppose n = 300 − 2 = 298
Let the frequency of the loaded unknown fork be n 1 and the loaded known
fork be n 2 :
300 − n 1 = 5
( 1 )
298 − n 2 = 9
( 2 )
Also frequency changes in both the forks are the same
∴ 300 − n 2 = 298 − n 1
or
n 2 − n 1 = 2
( 3 )
Subtracting (2) from (1)
n 2 − n 1 = −6
( 4 )
Obviously (3) and (4) are inconsistent.
Case (ii) Suppose n = 300 + 2 = 302
300 − n 1 = 5
( 5 )
302 − n 2 = 9
( 6 )
also 300 − n 2 = 302 − n 1
or
n 1 − n 2 = 2
( 7 )
Subtracting (5) from (6)
n 1 − n 2 = 2
( 8 )
Thus (7) and (8) are consistent. Therefore, correct solution is n = 302.