Unformatted text preview: e earth. We will refer to these two situations as A and B, respectively.
Since the proper time interval always has the same value, (∆t0 )A = (∆t0 )B . We can express
both sides of this expression using Equation 28.1. The result can be solved for ∆tB. SOLUTION Use of Equation 28.1 gives
2
2
∆tA 1 − vA / c 2 = ∆tB 1 − vB / c 2 ∆tB = ∆tA 2
1 − vA / c 2
2
1 − vB / c 2 = ∆tA 1 − (vA / c )2
1 − (vB / c )2 = (37.0 h) 1 – (0.75c/c )2
= 72 h
1 – (0.94c/c) 2 ______________________________________________________________________________
4. REASONING When you measure your breathing rate, you count N = 8.0 breaths during a
proper time interval of ∆t0 = 1.0 minutes, and in so doing you determine a rate of
R0 = N/∆t0 = (8.0 breaths)/(1.0 minute) = 8.0 breaths/minute. When measured by monitors
on the earth, the N = 8.0 breaths occur in a dilated time interval ∆t that is related to the Chapter 28 Problems ∆t0 proper time interval by ∆t = v2
1− 2
c
monitor on the earth, then, is given by (Equation 28.1). The breathing rate R measured by a R=
SOLUTION Substituting ∆t = 1479 ∆t 0
v2
1− 2
c N
∆t (1) (Equation 28.1) into Equation (1), we obtain (
)
v2
( 8.0 breaths ) 1 − 0.975c
c2 =
c2
= 1.8 breaths/minute
∆t0
1.0 minute
2 R= N
N
=
=
∆t ∆t0 2 1− v c2 N 1− 5. REASONING The observer is moving with respect to the oscillating object. Therefore, to
the observer, the oscillating object is moving with a speed of v = 1.90 × 108 m/s, and the
observer measures a dilated time interval for the period of oscillation. To determine this
dilated time interval ∆t = Tdilated, we must use the timedilation equation: ∆t = Tdilated = ∆t0 (28.1) v2
1− 2
c where ∆t0 is the proper time interval, as measured in the reference frame to which the fixed
end of the spring is attached. The proper time interval is the period T of the oscillation as
given by Equations 10.4 and 10.11: ∆t0 = T = 2π m
k where m is the mass of the object and k is the spring constant. (1) 1480 SPECIAL RELATIVITY SOLUTION Substituting Equation (1) into the timedilation equation gives
m
6.00 kg
2π
k=
76.0 N/m
=
= 2.28 s
2
2
8
v
(
)
1− 2
1 − 1.90 × 10 m/s 2
c
( 3.00 ×108 m/s )
2π Tdilated 6. REASONING The distance d traveled by the ship, according to an observer on earth, is
equal to the product of the speed v of the ship relative to earth and the elapsed time ∆t
measured by the earthbound observer, according to Equation 2.1:
d = v ∆t (2.1) The time interval ∆t is the dilated time interval and is related to the proper time interval ∆t0
for the journey (as measured by an observer on the ship) via Equation 28.1: ∆t = ∆t0 (28.1) v2
1− 2
c In this equation, v is the speed of the ship relative to earth, and c is the speed of light in a
vacuum. We will use Equation 28.1 to determine the speed v of the ship, and then
Equation 2.1 to find the distance the ship travels, according to the earthbound observer. SOLUTION Squaring both sides of Equation 28.1 and solving for the ratio v2/c2, we obtain ( ∆t ) 2 ( ∆t0 )
= 2 v2 ( ∆t0 )
1− 2 =
c
( ∆t )2 2 or v2
1− 2
c 2 or ∆t v2
1− 0 = 2 ∆t c (1) Solving Equation (1) for v yields ∆t 2 v = c 1− 0 ∆t 2 2 or ∆t v = c 1− 0 ∆t Substituting Equation (2) into Equation 2.1, we find that 2 (2) Chapter 28 Problems 1481 2 ∆t d = v∆t = c 1 − 0 ∆t ∆t 2 3.16 × 107 s 9.2 yr = 3.0 × 10 m/s 1 − 12 yr = 7.3 × 1016 m 12 yr 1 yr ______________________________________________________________________________ ( 7. 8 ) ( ) REASONING AND SOLUTION The proper time is the time it takes for the bacteria to
double its number, i.e., ∆t0 = 24.0 hours. For the earth based sample to grow to 256
bacteria, it would take 8 days (2n = 256 or n = 8). The "doubling time" for the space culture
would be
∆t0
24.0 h
∆t =
=
= 48.0 h or 2 days
2
2
v 0.866 c 1– 1– c
c In eight earth days, the space bacteria would undergo n' = ( 1 )8
2 = 4 "doublings". The number of space bacteria is
4 Number of Space Bacteria = 2n' = 2 = 16
______________________________________________________________________________
8. REASONING The tourist is moving at a speed of v = 1.3 m/s with respect to the path and,
therefore, measures a contracted length L instead of the proper length of L0 = 9.0 km. The
contracted length is given by the lengthcontraction equation, Equation 28.2.
SOLUTION According to the lengthcontraction equation, the tourist measures a length that
is (1.3 m/s )
v2
L = L0 1 − 2 = ( 9.0 km ) 1 −
= 8.1 km
c
( 3.0 m/s )2
2 9. SSM WWW REASONING All standard meter sticks at rest have a length of 1.00 m
for observers who are at rest with respect to them. Thus, 1.00 m is the proper length L0 of the meter stick. When the meter stick moves with speed v relative to an earthobserver, its
length L = 0.500 m will be a contracted length. Since both L0 and L are known, v can be ( ) found directly from Equation 28.2, L = L0 1 – v 2 / c 2 . 1...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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