Physics 295 Solutions to problems from Chapter 18
P18.1 y  y1  y2  3.00cos 4.00x  1.60t   4.00sin  5.0x  2.00t  evaluated at the given x
values.
(a)
x  1.00 , t  1.00
y  3.00cos 2.40 rad   4.00sin  3.00 rad   1.65 cm
(b)
x  1.00 , t  0.500
y  3.00cos 3.20 rad   4.00sin  4.00 rad   6.02 cm
(c)
x  0.500 , t  0
y  3.00cos 2.00 rad   4.00sin  2.50 rad   1.15 cm
P18.12 From y  2A0 sin kx cost we find
y
 2A0 k cos kx cost
x
y
 2A0 sin kx sin t
t
2 y
 2A0 k2 sin kx cost
x2
2 y
 2A0 2 sin kx cost
t 2
Substitution into the wave equation
gives
 1
2A0 k2 sin kx cost   2   2A0 2 sin kx cost 
v 
This is satisfied, provided that v 


. But this is true, because v   f  2 f 
2
k
P18.15 y1  3.00sin   x  0.600t  cm ; y2  3.00sin   x  0.600t  cm
y  y1  y2  3.00sin  x  cos 0.600 t   3.00sin  x  cos 0.600 t  cm
y   6.00 cm  sin  x  cos 0.600 t 
(a)
We can take cos 0.600 t   1 to get the maximum y.
At x  0.250 cm ,
ymax   6.00 cm  sin  0.250   4.24 cm
(b)
At x  0.500 cm ,
ymax   6.00 cm  sin  0.500   6.00 cm
(c)
Now take cos 0.600 t   1 to get ymax :
At x  1.50 cm ,
ymax   6.00 cm  sin 1.50  1  6.00 cm

k
(d)
The antinodes occur when
But k 
2

x
n
 n  1, 3, 5,
4
  2.00 cm
  so
x1 
and
P18.17 L  30.0 m ;   9.00  103


 0.500 cm as in (b)
4
3
x2 
 1.50 cm as in (c)
4
5
x3 
 2.50 cm
4
v
kg m ; T  20.0 N ; f1 
2L
12
where
T 
v 

so
f1 
 47.1 m s
47.1
 0.786 Hz
60.0
f2  2 f1  1.57 Hz
f3  3 f1  2.36 Hz
f 4  4 f1  3.14 Hz

P18.20 For the whole string vibrating, dNN  0.64 m  ;   1.28 m The
2
1
speed of a pulse on the string is v  f   330 1.28 m  422 m s
s
2

(a)
When the string is stopped at the fret, dNN  0.64 m  ;
3
2
  0.853 m
v 422 m s
f  
 495 Hz
 0.853 m
(b)
The light touch at a point one third of the way along the
string damps out vibration in the two lowest vibration
states of the string as a whole. The whole string vibrates in
its third resonance possibility:

  0.427 m
2
v 422 m s
f  
 990 Hz
 0.427 m
3dNN  0.64 m  3
P18.31
FIG. P18.20(a)
The wavelength is  
FIG. P18.20(b)
v 343 m s

 1.31 m
f
261.6 s
Fig. P18.31
so the length of the open pipe vibrating in its simplest
(A-N-A) mode is
1
dA to A    0.656 m
2
A closed pipe has (N-A) for its simplest resonance,
(N-A-N-A) for the second,
and (N-A-N-A-N-A) for the third.
Here, the pipe length is
5dN to A 
5 5
 1.31 m   1.64 m
4 4
P18.37 For resonance in a narrow tube open at one end,
f n
(a)
v
 n  1, 3, 5,
4L
f = 384 Hz

Assuming n  1 and n  3 ,
v
3v
and 384 
384 
4  0.228
4  0.683
warm
air
22.8 cm
68.3 cm
In either case, v  350 m s .
(b)
For the next resonance n  5 , and L 
5v 5  350 m s

 1.14 m
4f
4  384 s1 
FIG. P18.37
P18.43 f  v  T
f new  110
f  110/ s  104.4/ s = 5.64 beats s
540
 104.4 Hz
600