Unformatted text preview: yields v JW = v JS − v WS . We assume that both
the jetskier and the waves move in the positive direction. Since the jetskier is traveling
faster than the waves, vJS (the magnitude of vJS) is greater than vWS (the magnitude of vWS).
Therefore, our vector result v JW = v JS − v WS becomes vJW = vJS − vWS , which is an
expression relating the speeds. Substituting this result into Equation (2), we obtain vJW = vJS − vWS = f λ (3) Solving Equation (3) for the wave speed vWS, we obtain vWS = vJS − f λ = 8.4 m/s − (1.2 Hz ) ( 5.8 m ) = 1.4 m/s 9. REASONING During each cycle of the wave, a particle of the string moves through a total
distance that equals 4A, where A is the amplitude of the wave. The number of wave cycles
per second is the frequency f of the wave. Therefore, the distance moved per second by a
string particle is 4Af. The time to move through a total distance L, then, is t = L/(4Af).
According to Equation 16.1, however, we have that f = v/λ, so that t= L
4 Af 4 A ( v / λ ) SOLUTION Using the result obtained above, we find that (
( ) 1.0 × 103 m ( 0.18 m )
= 5.0 × 101 s
4 Av 4 2.0 × 10 –3 m ( 450 m/s ) ) ______________________________________________________________________________ 838 WAVES AND SOUND 10. REASONING AND SOLUTION First find the speed of the record at a distance of 0.100 m
from the center:
v = rω = (0.100 m)(3.49 rad/s) = 0.349 m/s
The wavelength is, then, λ = v/f = (0.349 m/s)/(5.00 × 103 Hz) = 6.98 × 10 –5 m (16.1) ______________________________________________________________________________
11. REASONING AND SOLUTION When traveling with the waves, the skier "sees" the
waves to be traveling with a velocity of vw – vs and with a period of T1 = 0.600 s.
Suppressing the units for convenience, we find that the distance between crests is (in
meters) λ = (vs – vw)T1 = (12.0 − vw)(0.600) = 7.20 – 0.600 vw (1) Similarly, for the skier traveling opposite the waves λ = (vs + vw)T2 = 6.00 + 0.500 vw (2) a. Subtracting Equation (2) from Equation(1) and solving for vw gives vW = 1.09 m/s .
b. Substituting the value for vw into Equation (1) gives λ = 6.55 m .
12. REASONING The speed v of a transverse wave on a wire is given by v = F / ( m /L )
(Equation 16.2), where F is the tension and m/L is the mass per unit length (or linear
density) of the wire. We are given that F and m are the same for the two wires, and that one
is twice as long as the other. This information, along with knowledge of the wave speed on
the shorter wire, will allow us to determine the speed of the wave on the longer wire.
SOLUTION The speeds on the longer and shorter wires are:
[Longer wire] vlonger = F
m /L longer [Shorter wire] vshorter = F
m /L shorter Dividing the expression for vlonger by that for vshorter gives Chapter 16 Problems vlonger
m /L longer = F
m /L shorter = 839 L longer
L shorter Noting that vshorter = 240 m/s and that Llonger = 2Lshorter, the speed of the wave on the longer
vlonger = vshorter
= ( 240 m/s )
= ( 240 m/s ) 2 = 340 m/s
13. SSM REASONING The tension F in the violin string can be found by solving Equation 16.2 for F to obtain F = mv 2 / L , where v is the speed of waves on the string and
can be found from Equation 16.1 as v = f λ .
SOLUTION Combining Equations 16.2 and 16.1 and using the given data, we obtain ( ) ( ) 2
= (m / L) f 2λ 2 = 7.8 ×10−4 kg/m ( 440 Hz ) 65 ×10−2 m = 64 N
______________________________________________________________________________ F= 14. REASONING The length L of the string is one of the factors that affects the speed of a
wave traveling on it, in so far as the speed v depends on the mass per unit length m/L
according to v =
(Equation 16.2). The other factor affecting the speed is the tension
F. The speed is not directly given here. However, the frequency f and the wavelength λ are
given, and the speed is related to them according to v = f λ (Equation 16.1). Substituting
Equation 16.1 into Equation 16.2 will give us an equation that can be solved for the length
SOLUTION Substituting Equation 16.1 into Equation 16.2 gives
v= fλ = F
m/ L Solving for the length L, we find that ( ) −3
f 2λ 2 m ( 260 Hz ) ( 0.60 m ) 5.0 × 10 kg
= 0.68 m
2 2 840 WAVES AND SOUND 15. REASONING The speed v of the transverse pulse on the wire is determined by the tension
F in the wire and the mass per unit length m/L of the wire, according to v =
(Equation 16.2). The ball has a mass M. Since the wire supports the weight Mg of the ball
and since the weight of the wire is negligible, it is only the ball’s weight that determines the
tension in the wire, F = Mg. Therefore, we can use Equation 16.2 with this value of the
tension and solve it for the acceleration g due to gravity. The speed of the transverse pulse
is not given, but we know that the pulse travels the length L of the wire in a time t and that
the speed is v = L/t.
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