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the wire’s linear density.
SOLUTION According to Equation 17.3, the harmonic number n is
f 2L
(1)
n= n
v
F
(Equation 16.2), where F is the tension and m/L is the linear
The speed v is v =
m/ L
density. Substituting this expression into Equation (1) gives n= fn 2 L
v = fn 2L
F
m/ L = fn 2L m/ L
0.0140 kg/m
= ( 20.0 Hz ) 2 ( 7.60 m )
=2
F
323 N 35. CONCEPT QUESTIONS The fundamental frequency f1 of the wire is given by f1 = v/(2L)
(Equation 17.3, with n = 1), where v is the speed at which the waves travel on the wire and L
is the length of the wire. The speed is related to the tension F in the wire according to
F
v=
(Equation 16.2), where m/L is the mass per unit length of the wire.
m/L 924 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA The tension in the wire in Part 2 of the drawing is less than the tension in Part 1. The reason
is related to Archimedes’ principle (see Equation 11.6). This principle indicates that when
an object is immersed in a fluid, the fluid exerts an upward buoyant force on the object. In
Part 2 the upward buoyant force from the mercury supports part of the block’s weight, thus
reducing the amount of the weight that the wire must support.
SOLUTION Substituting Equation 16.2 into Equation 17.3, we can obtain the fundamental
frequency of the wire:
v
1
F
f1 =
=
(1)
2L 2L m / L In Part 1 of the drawing, the tension F balances the weight of the block, keeping it from
falling. The weight of the block is its mass times the acceleration due to gravity (see
Equation 4.5). The mass, according to Equation 11.1 is the density ρcopper times the volume
V of the block. Thus, the tension in Part 1 is
Part 1 tension F = ( mass ) g = ρcopperVg In Part 2 of the drawing, the tension is reduced from this amount by the amount of the
upward buoyant force. According to Archimedes’ principle, the buoyant force is the weight
of the liquid mercury displaced by the block. Since half of the block’s volume is immersed,
the volume of mercury displaced is 1 V . The weight of this mercury is the mass times the
2
acceleration due to gravity. Once again, according to Equation 11.1, the mass is the density
ρmercury times the volume, which is 1 V . Thus, the tension in Part 2 is
2 Part 2 tension F = ρcopperVg − ρ mercury ( 1V ) g
2 With these two values for the tension we can apply Equation (1) to both parts of the drawing
and obtain
f1 = Part 1 Part 2 1
f1 =
2L 1
2L ρ copperVg
m/L ρ copperVg − ρ mercury ( 1 V ) g
2
m/L Dividing the fundamental frequency of Part 2 by that of Part 1 gives Chapter 17 Problems f1, Part 2
f1, Part 1 1 ρ copperVg − ρ mercury
m/L
= 2L
1 ρ copperVg
2L
m/L = (1V ) g
2 ( 8890 kg/m 3 − 1 13 600 kg/m3
2
8890 kg/m 3 = )= 925 ρ copper − 1 ρ mercury
2
ρ copper 0.485 36. REASONING Each string has a node at each end, so the frequency of vibration is given by
Equation 17.3 as fn = nv/(2L), where n = 1, 2, 3, … The speed v of the wave can be
determined from Equation 16.2 as v = F / ( m / L ) . We will use these two relations to find
the lowest frequency that permits standing waves in both strings with a node at the junction.
SOLUTION
Since the frequency of the left string is equal to the frequency of the right string, we can
write
F
F
nright
nleft
( m / L )right
( m / L )left
=
2Lleft
2L right Substituting in the data given in the problem yields nleft 190.0 N
190.0 N
nright
−2
6.00 × 10 kg/m
1.50 × 10−2 kg/m
=
2 ( 3.75 m )
2 (1.25 m ) This expression gives nleft = 6nright. Letting nleft = 6 and nright = 1, the frequency of the left
string (which is also equal to the frequency of the right string) is (6)
f6 = 190.0 N
6.00 × 10−2 kg/m
=
2 ( 3.75 m ) 45.0 Hz 37. SSM REASONING We can find the extra length that the Dtuner adds to the Estring by
calculating the length of the Dstring and then subtracting from it the length of the E string.
For standing waves on a string that is fixed at both ends, Equation 17.3 gives the frequencies 926 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA as f n = n(v / 2 L) . The ratio of the fundamental frequency of the Dstring to that of the
Estring is
f D v /(2 LD ) LE
=
=
f E v /(2 LE ) LD This expression can be solved for the length LD of the Dstring in terms of quantities given
in the problem statement.
SOLUTION The length of the Dstring is
f 41.2 Hz LD = LE E = (0.628 m) = 0.705 m
f 36.7 Hz D
The length of the Estring is extended by the Dtuner by an amount LD − LE = 0.705 m − 0.628 m = 0.077 m 38. REASONING The beat frequency is equal to the higher frequency of the shorter string
minus the lower frequency of the longer string. The reason the longer string has the lower
frequency can be seen from the drawing, where it is evident that both strings are vibrating at
their fundamental frequencies. The fundamental frequency of vibration (n = 1) for a string
fixed at each end is given by f1 = v/(2L) (Equation 17.3). Since the speed v is the same for
both strings (see the following paragraph), but the length L is greater for the longer string,
the longer...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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