Physics II
Homework IV
CJ
Chapter 13; 68, 96
Chapter 15; 4, 11, 26, 30, 41, 50, 74
13.68. IDENTIFY:
In SHM,
amax 
k
A.
mtot
Apply
 F = ma to the top block.
SET UP: The maximum acceleration of the lower block can’t exceed the maximum
acceleration that can be given to the other block by the friction force.
EXECUTE: For block m, the maximum friction force is fs  s n  s mg .  Fx  max gives
s mg  ma and a  s g . Then treat both blocks together and consider their simple harmonic
motion.
 k 
amax  
A.
M m
Set
amax  a
and solve for A: s g  
k 
A
M m
and A  s g (M  m) .
k
EVALUATE: If A is larger than this the spring gives the block with mass M a larger
acceleration than friction can give the other block, and the first block accelerates out from
underneath the other block.
13.96. IDENTIFY:
SET UP:
Calculate
Fnet and
define
keff by Fnet  keff x . T  2 m / keff
If the elongations of the springs are
x1 and x2 ,
.
they must satisfy
x1  x2  0.200 m
EXECUTE: (a) The net force on the block at equilibrium is zero, and so k1x1  k2 x2 and one
spring (the one with k1  2.00 N/m ) must be stretched three times as much as the one with
k2  6.00  m . The sum of the elongations is 0.200 m, and so one spring stretches 0.150 m
and the other stretches 0.050 m, and so the equilibrium lengths are 0.350 m and 0.250 m.
(b) When the block is displaced a distance x to the right, the net force on the block is
k1  x1  x   k2  x2  x   k1x1  k2 x2    k1  k2  x. From the result of part (a), the term in square
brackets is zero, and so the net force is   k1  k2  x, the effective spring constant is keff  k1  k2
and the period of vibration is
T  2
0.100 kg
 0.702 s.
8.00  m
EVALUATE: The motion is the same as if the block were attached to a single spring that
has force constant keff .
15.4. IDENTIFY:
SET UP:
EXECUTE:
f  v
1.0 mm  0.0010 m
f 
v


1500 m s
 1.5 106 Hz
0.0010 m
EVALUATE: The frequency is much higher than the upper range of human hearing.
15.11. IDENTIFY and SET UP: Read A and T from the graph. Apply Eq.(15.4) to determine  and
then use Eq.(15.1) to calculate v.
EXECUTE:
(a) The maximum y is 4 mm (read from graph).
(b) For either x the time for one full cycle is 0.040 s; this is the period.
and t  0 and since the wave is traveling in the  x-direction then
y( x, t )  Asin[2 (t / T  x / )]. (The phase is different from the wave described by Eq.(15.4); for
that wave y  A for x  0, t  0.) From the graph, if the wave is traveling in the  x-direction
and if x  0 and x  0.090 m are within one wavelength the peak at t  0.01 s for x  0 moves so
that it occurs at t  0.035 s (read from graph so is approximate) for x  0.090 m. The peak for
x  0 is the first peak past t  0 so corresponds to the first maximum in sin[2 (t / T  x /  )] and
hence occurs at 2 (t / T  x / )   / 2. If this same peak moves to t1  0.035 s at x1  0.090 m, then
(c) Since y  0 for
x0
2 (t1 / T  x1 /  )   / 2
Solve for  :
t1 / T  x1 /   1/ 4
x1 /   t1 / T  1/ 4  0.035 s / 0.040 s  0.25  0.625
  x1 / 0.625  0.090 m/ 0.625  0.14 m.
Then v  f    / T  0.14 m/ 0.040 s  3.5 m/s.
(d) If the wave is traveling in the  x-direction, then y( x, t )  Asin(2 (t / T  x / )) and the peak at
t  0.050 s for x  0 corresponds to the peak at t1  0.035 s for x1  0.090 m. This peak at x  0 is
the second peak past the origin so corresponds to 2 (t / T  x / )  5 / 2. If this same peak
moves to t1  0.035 s for x1  0.090 m, then 2 (t1 / T  x1 /  )  5 / 2.
t1 / T  x1 /   5/ 4
x1 /   5/ 4  t1 / T  5/ 4  0.035 s / 0.040 s  0.375
  x1 / 0.375  0.090 m/ 0.375  0.24 m.
Then v  f    / T  0.24 m/ 0.040 s  6.0 m/s.
EVALUATE: No. Wouldn’t know which point in the wave at
x0
moved to which point at
x  0.090 m.
15.26. IDENTIFY: The distance the wave shape travels in time t is vt. The wave pulse reflects at
the end of the string, at point O.
SET UP: The reflected pulse is inverted when O is a fixed end and is not inverted when O
is a free end.
EXECUTE: (a) The wave form for the given times, respectively, is shown in Figure 15.26a.
(b) The wave form for the given times, respectively, is shown in Figure 15.26b.
EVALUATE: For the fixed end the result of the reflection is an inverted pulse traveling to
the left and for the free end the result is an upright pulse traveling to the left.
Figure 15.26
15.30. IDENTIFY: Apply the principle of superposition.
SET UP: The net displacement is the algebraic sum of the displacements due to each pulse.
EXECUTE: The shape of the string at each specified time is shown in Figure 15.30.
EVALUATE: The pulses interfere when they overlap but resume their original shape after
they have completely passed through each other.
Figure 15.30
15.41. IDENTIFY: Compare y( x, t ) given in the problem to Eq.(15.28). From the frequency
and wavelength for the third harmonic find these values for the eighth harmonic.
(a) SET UP: The third harmonic standing wave pattern is sketched in Figure 15.41.
Figure 15.41
EXECUTE: (b) Eq. (15.28) gives the general equation for a standing wave on a
string:
y( x, t )  ( ASW sin kx)sin t
ASW  2 A,
so
A  ASW / 2  (5.60 cm)/2  2.80 cm
(c) The sketch in part (a) shows that L  3( / 2). k  2 /  ,   2 / k
Comparison of y( x, t ) given in the problem to Eq. (15.28) gives k  0.0340 rad/cm. So,
  2 /(0.0340 rad/cm)  184.8 cm
L  3( / 2)  277 cm
(d)   185 cm, from part (c)
  50.0 rad/s so f   / 2  7.96 Hz
period T  1/ f  0.126 s
v  f   1470 cm/s
(e)
v y  dy / dt   ASW sin kx cos t
v y , max   ASW  (50.0 rad/s)(5.60 cm)  280 cm/s
(f )
f3  7.96 Hz  3 f1,
so
f1  2.65 Hz
is the fundamental
f8  8 f1  21.2 Hz; 8  2 f8  133 rad/s
  v / f  (1470 cm/s)/(21.2 Hz)  69.3 cm and k  2 /   0.0906 rad/cm
y( x, t )  (5.60 cm)sin([0.0906 rad/cm]x)sin([133 rad/s]t )
EVALUATE: The wavelength and frequency of the standing wave equals the wavelength
and frequency of the two traveling waves that combine to form the standing wave. In the 8th
harmonic the frequency and wave number are larger than in the 3rd harmonic.
15.50. IDENTIFY: Compare y( x, t ) given in the problem to the general form given in Eq.(15.8).
SET UP: The comparison gives
EXECUTE: (a)
A  0.750 cm,  
A  0.750 cm , k  0.400 rad/cm and   250 rad/s .
2
 5.00 cm, f  125 Hz, T  1  0.00800 s
0.400 rad/cm
f
and
v   f  6.25 m/s.
(b) The sketches of the shape of the rope at each time are given in Figure 15.50.
(c) To stay with a wavefront as t increases,
 x -direction.
x decreases
and so the wave is moving in the
(d) From Eq. (15.13), the tension is
F   v 2  (0.50 kg m) (6.25 m s) 2  19.5 N .
(e) Pav  12  F 2 A2  54.2 W.
EVALUATE: The argument of the cosine is (kx  t ) for a wave traveling in the
and that is the case here.
 x-direction ,
Figure 15.50
15.74. IDENTIFY: The standing wave is given by Eq.(15.28).
SET UP: At an antinode,
sin kx  1 . v y ,max   A . ay ,max   2 A .
EXECUTE: (a)   v f  (192.0 m s) (240.0 Hz)  0.800 m , and the wave amplitude is
ASW  0.400 cm. The amplitude of the motion at the given points is
(i) (0.400 cm)sin ( )  0 (a node) (ii) (0.400 cm) sin( 2)  0.400 cm (an antinode)
(iii) (0.400 cm) sin( 4)  0.283 cm
(b) The time is half of the period, or 1 (2 f )  2.08 103 s.
(c) In each case, the maximum velocity is the amplitude multiplied by   2 f and the
maximum acceleration is the amplitude multiplied by  2  4 2 f 2 :
(i) 0, 0; (ii) 6.03 m s, 9.10  103 m s 2 ; (iii) 4.27 m s, 6.43  103 m s 2 .
EVALUATE: The amplitude, maximum transverse velocity, and maximum transverse
acceleration vary along the length of the string. But the period of the simple harmonic
motion of particles of the string is the same at all points on the string.