Page 196 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 15] WAVES AND SOUND 181

9
6
(a) Since 1 MHz = 10 Hz, 9400 MHz = 9.4 × 10 Hz, and
8
c 3.00 × 10 m/s
λ = = = 3.19 × 10 −2 m = 3.19 cm
9
f 9.4 × 10 Hz
(b) Since 1 µs = 10 −6 s, 0.08 µs = 8 × 10 −8 s, and the length s of each wave group is
8
s = ct = (3.00 × 10 m/s)(8 × 10 −8 s) = 24 m
(c) There are two ways to find the number n of waves in each group:
9
n = ft = (9.4 × 10 Hz)(8 × 10 −8 s) = 752 waves
s 24 m
n = = = 752 waves
λ 3.19 × 10 −2 m

SOLVED PROBLEM 15.5
An anchored boat is observed to rise and fall through a total range of 2 m once every 4 s as waves whose
crests are 30 m apart pass it. Find (a) the frequency of the waves, (b) their velocity, (c) their amplitude,
and (d) the velocity of an individual water particle at the surface.
1 1
(a) f = = = 0.25 Hz
T 4s
(b) v = f λ = (0.25 Hz)(30 m) = 7.5 m/s
(c) The amplitude is half the total range so A = 1m.
(d) As each wave passes, the water particles at the surface move in circular orbits of radius r = A = 1 m (see
Fig. 15-3). The circumference of such an orbit is

s = 2πr = (2π)(1m) = 6.28 m
The waves have the period 4 s, which means that each surface water particle must move through its 6.28-m
orbit in 4 s. The velocity of such a water particle is therefore
s 6.28 m
V = = = 1.57 m/s
T 4s
Note that the wave velocity here is 7.5 m/s, nearly five times greater. This signifies that the motion of a wave
can be much faster than the motions of the individual particles of the medium in which the wave travels.

LOGARITHMS
Although logarithms have many other uses, their chief application in applied physics is in connection with the
decibel, which is described in the next section. Logarithms are discussed here only to the extent required for this
purpose.
n
The logarithm of a number N is the power n to which 10 must be raised in order that 10 = N. That is,
N = 10 n therefore log N = n

(Logarithms are not limited to a base of 10, but base-10 logarithms are the most common and are all that are
needed here.) For instance,
1000 = 10 3 therefore log 1000 = 3
0.01 = 10 −2 therefore log 0.01 =−2