Physics Solution Manual for 1100 and 2101

# In example 6 it is shown that when the source and the

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Unformatted text preview: s of zero field as λ = 2d (1) SOLUTION Solving c = f λ (Equation 16.1) for f yields f= c (2) λ Substituting Equation (1) into Equation (2), we obtain f = 9. SSM WWW c = λ c 3.00 × 108 m/s = = 2d 2 ( 0.34 m ) 4.4 × 108 Hz REASONING AND SOLUTION According to Equation 16.1, the wavelength is λ = c/f. Since the rods can be adjusted so that each one has a length of λ/4, the length of each rod is ( )( ) 3.00 ×10 m/s / 60 ×10 Hz c/ f L= = = = 1.25 m 4 4 4 ______________________________________________________________________________ λ 8 6 10. REASONING AND SOLUTION The VHF wavelength is = λ c 3.00 ×108 m/s = = 4.76 m f 63.0 ×106 Hz = λ c 3.00 ×108 m/s = = 0.569 m f 527 ×106 Hz The UHF wavelength is The ratio of the wavelengths is (4.76 m)/(0.569 m) = 8.37 1284 ELECTROMAGNETIC WAVES ______________________________________________________________________________ 11. SSM REASONING AND SOLUTION The number of wavelengths that can fit across the width W of your thumb is W/λ. From Equation 16.1, we know that λ = c/f, so W Wf (2.0 × 10−2 m)(5.5 × 1014 Hz) == = 3.7 × 104 8 c λ 3.0 × 10 m/s ______________________________________________________________________________ No. of wavelengths = 12. REASONING The length of each pulse is equal to the product of its speed and the time, or x = ct0. According to Equation 16.1, the wavelength λ in a vacuum is related to the frequency f by λ = c / f , where c is the speed of light in a vacuum. When the light travels in water, the speed is no longer c, but its value v is given. The wavelength in water, then, is λ = v / f . The frequency in a vacuum and in water is the same. SOLUTION a. The number of wavelengths in one pulse is equal to the length of the pulse divided by the wavelength. The length of each pulse is x = ct0 and the wavelength is λ = c / f , so Number of wavelengths = x λ = ct0 c/ f ( )( ) = = × 1014 Hz 2.7 × 10−11 s = × 104 f t0 5.2 1.4 b. When the light is traveling in water, its speed is v, which is less than the speed of light in a vacuum. The length of each pulse is now x = vt0 and the wavelength is λ = v / f , so Number of wavelengths = x λ = vt0 v/ f ( )( ) = = × 1014 Hz 2.7 × 10−11 s = × 104 f t0 5.2 1.4 ______________________________________________________________________________ 13. REASONING To determine the difference in frequencies, we will calculate each frequency and subtract one from the other. Each frequency f is related to the wavelength λ and the speed of light c according to f = c/ λ (Equation 16.1). SOLUTION Using Equation 16.1 to calculate each frequency, we find that f 2 − f1 = c λ2 − ( ) 1 1 7 = 2.9979 × 108 m/s − = 4.500 ×10 Hz λ1 0.34339 m 0.36205 m c Chapter 24 Problems 1285 k (Equation 10.11), the angular oscillation frequency m ω depends upon the mass m of the oscillator and the spring constant k. The angular frequency ω (in rad/s) of the motion of the oscillating mass is related to the frequency f (in Hz) of the resulting ELF radio waves by ω = 2π f (Equation 10.6). We will use c = f λ (Equation 16.1) to determine the frequency of the ELF radio waves, where c is the speed of light in a vacuum and λ is the wavelength. 14. REASONING According to ω = SOLUTION Squaring both sides of ω = k (Equation 10.11) and solving for k, we obtain m k = ω2 = mω 2 or k m (1) Substituting ω = 2π f (Equation 10.6) into Equation (1) yields k = 4π 2 mf 2 = m ( 2π f ) 2 Solving c = f λ (Equation 16.1) for f gives f = c λ (2) . Substituting this result into Equation (2), we find that 2 3.00 ×108 m/s c k = 4= 4π 2 ( 0.115 kg ) = 177 N/m π m 4.80 ×107 m λ ______________________________________________________________________________ 2 2 15. SSM REASONING We proceed by first finding the time t for sound waves to travel between the astronauts. Since this is the same time it takes for the electromagnetic waves to travel to earth, the distance between earth and the spaceship is dearth -ship = ct . SOLUTION The time it takes for sound waves to travel at 343 m/s through the air between the astronauts is d 1.5 m –3 t = astronaut = = 4.4 × 10 s 343 m/s vsound Therefore, the distance between the earth and the spaceship is 8 –3 6 dearth -ship = ct = (3.0 × 10 m/s)(4.4 × 10 s) = 1.3 × 10 m ______________________________________________________________________________ 1286 ELECTROMAGNETIC WAVES 16. REASONING The mirror must rotate with a sufficient angular speed ω so that, after reflecting light from the source toward the fixed mirror, one of its faces is in the correct position to intercept light returning from the fixed mirror and reflect it toward the observer. Since the rotating mirror in Michelson’s setup has eight sides, the minimum angular displacement Δθ meeting this condition is one eighth of a revolution: Δθ = 0.125 rev. The mirror must rotate at least this far in the time Δt it takes the light to travel to the fixed mirror and back. The minimum, constant, angular speed of the mirror, then, can be found from ∆θ (Eq...
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