8.3 Solutions
387
K i S i = 10 × 18 × 0.6
+ 10 × 18 × 0.02 + (18 × 4 + 10 × 4) × 2 × 0.03 + 50 × 0.5
= 143.3
t R =
0.16 × 720
143.3
= 0.8 s
8.3.8 Echo
8.79 The drum rate, that is, frequency when the echo is inaudible, is 40/min or
2/3 per second. Therefore, the time period of drum beats t 1 =
3
2 s. Time for
the echo, t 1 =
2x
v , where x is the initial distance from the mountain and v is
the sound velocity.
Thus
2x
v
=
3
2
(1)
On moving 90 m towards the mountain ht is x − 90 m from the mountain, the
drum rate is 60/min or 1/s and again the echo is not heard.
Thus
2(x − 90)
v
= 1
( 2 )
Solving (1) and (2) we get
x = 270 m and v = 360 m/s.
8.80 (a) If the width of the valley is d m and the rifle shot is fired at a distance x
from one of the mountains the echoes will be heard in time t 1 and t 2 s:
t 1 =
2x
v
(1)
t 2 = 2
(d − x)
v
(2)
Adding (1) and (2)
t 1 + t 2 = 2 + 4 =
2d
v
=
2d
360
Therefore d = 1080 m.
(b) Solving (1) with t 1 = 2 s and v = 360 m we find x = 360 m and therefore
d − x = 720 m. Subsequent echoes will be heard after 6, 8, 10,. . . s.