Physics 2503 Test #4 - Solutions

Physics 2503 – Test #4 - Solutions

1. Two waves travel simultaneously through the same medium. The first wave is

described by y₁(x,t) = (2.00 cm)cos(kx-ωt) and the second wave by y₂(x,t) = (3.00

cm)sin(kx-ωt). What is the amplitude of the resulting wave?

(a) 1.00 cm (b) 2.50 cm (c) 3.61 cm (d) 5.00 cm (e) 3.94 cm

Ans: (c) The superposition of two traveling waves is also a traveling wave of the same frequency and wavelength, but with different amplitude and a phase shift.

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1 2 cos( ) sin( )

cos( ) cos( ) cos sin( )sin

y y y A kx t B kx t y C kx t C kx t kx t ω ω ω δ ω δ ω δ = + = − + − = − + = − − −

For both of these equations to hold….

cos and sin

C δ = A −C δ =B

Square these, add, and take a square root: C = A2+B2 =3.61 cm.

2. Sound of frequency 600. Hz is emitted by a point source. The sound can travel along two different paths to reach an observer. What is the minimum non-zero path length difference that results in constructive interference? Assume the speed of sound in air is 340 m/s. (a) 28.3 cm (b) 1.13 cm (c) 1.76 cm (d) 88.2 cm (e) 56.7 cm

Ans: (e) Constructive interference requires a path difference of nλ, and the “minimum non-zero path length” corresponds to n=1:

Path-length difference = λ = v/f = (340 m/s)/(600 s-1) = 0.567 m.

3. The left-hand diagram is a displacement graph at time t = 0 s showing two pulses on a rope that are approaching each other with speeds of 1.0 cm/s. A student uses the superposition principle to predict that the rope displacement at t = 3.0 s will be as shown in the right-hand graph. Which of the statements written below is accurate?

(a) The superposition principle was incorrectly applied to obtain the right-hand graph. (b) The total energy (kinetic plus potential) of each pulse in the left-hand graph is zero. (c) The total energy of the two pulses in the left-hand graph is zero because one has

positive displacement and the other has negative displacement. (d) All of the rope’s energy is kinetic at the time of the right-hand graph.

(e) The failure of conservation of energy demonstrates that the superposition principle is not valid for this system.

Ans: (d) All are nonsense except for (d). The energy has not gone away. The distortion is instantaneously zero, so there is no potential energy. All energy must all be kinetic and a moment later we’ll see distortion again.

4. A periodic wave is the superposition of harmonic waves of frequency 200 Hz, 300 Hz, and 500 Hz. What is the period of this wave?

(a) 0.667 ms (b) 0.333 ms (c) 2.00 ms (d) 10.0 ms (e) 5.00 ms

Ans: (d) The general form of the Fourier series expansion is

1 ( ) _nsin( _n) n F t A n tω φ ∞ =

=

∑

For a series with non-zero terms of frequency 200 Hz, 300 Hz and 500 Hz, the largest possible fundamental frequency is f = 100 Hz. The corresponding period is 1/f = 0.01 s = 10.0 ms, which is the only possible answer among the choices given.

5. A piano tuner hears an intensity variation of 5.0 Hz when a 440. Hz calibration tone is played at the same time as a piano wire that should have a (fundamental) standing wave frequency of 440. Hz. She has determined that the piano wire’s frequency is less than 440 Hz. What change in the tension of the piano wire is required to bring it into tune?

(a) -0.57% (b) -2.3% (c) -1.1% (d) +1.1% (e) +2.3%

Ans: (e) The frequency of intensity pulses is twice the beat frequency: f = 2fbeat =

f1-f2. In this case we have f1 = 440.0 Hz and f = 5.00 Hz, so f2 = 435.00 Hz.

For a string or wire, v T μ

= , and (with the linear mass density μ

constant) we adjust the tension T so that the standing wave frequency of the wire is f1 instead of f2.

1 2 1 2 1 1 2 2 2 ₂ 1 1 1 1 1 2 2 2 2 2 440 Hz 1.0231 435 Hz T T v v v f v f f v T T V f v T T V μ μ λ λ = = = = ⎛ ⎞ ⎛ ⎞ = = ⇒ =_{⎜ ⎟} =_{⎜ ⎟} = ⎝ ⎠ ⎝ ⎠

The tension must be 1.0231 times its initial value, so the change is +2.3%.

6. The magnetic field of a harmonic electromagnetic wave has a peak value of

10

5.0 10 T× − . What is the intensity of the wave?

(a) 13 2 W 1.0 10 m − × (b) 5 2 W 1.5 10 m − × (c) 5 2 W 3.0 10 m − × (d) 13 2 W 2.0 10 m − × (e) 5 2 W 7.5 10 m × Ans: (c) 2 ₂ 5 max 2 0 0 0 W 2.98 10 2 m c B _cB E B E cB S I S μ μ μ − × = = = = = = × G G G

7. The solar intensity at the orbit of the planet Venus is 2650 W/m2. What volume of space in this region contains 1.0 J of energy?

(a) 8.7×10-6 m3 (b) 1.7×102 m3 (c) 1.4×10-3 m3 (d) 0.73×103 m3 (e) 1.1×105 m3 Ans: (e) 2 6 5 3 3 W 2650 m J 1 J 8.833 10 1.13 10 m m I S I I uc u V c u − = = = ⇒ = = × ⇒ = = ×

8. The intensity of solar radiation near the earth is 3

2

W 1.4 10

m

× . What force is exerted

by solar radiation incident normally on a 5.0 m2 perfectly reflecting panel of an

artificial satellite orbiting the earth?

(a) 1.4×104 N (b) 9.4×10-5 N (c) 1.4×10-4 N (d) 2.3×10-5 N (e) 4.7×10-5 N Ans: (e) 2 2 5 W 1400 5m m 2 2 4.67 10 N I S A I F uA A c − = = = = = = ×

The factor of 2 is for “perfectly reflecting.”

9. Unpolarized light is incident on the surface of a pool of water (index of refraction 1.33). For what angle of incidence will the reflected light be completely linearly polarized? (a) 53° (b) 41° (c) 49° (d) 37° (e) 45°

Ans: (a) Polarization on reflection is complete when the angle of incidence is equal to Brewster’s angle (reflected and refracted rays perpendicular). tan θB = n = 1.33 ⇒ θB = 53°.

10. A laser produces a beam of wavelength 540 nm. If the output power is 10 mW, how many photons are emitted by the laser in one second?

(a) 6.9×1012 (b) 2.7×1016 (c) 3.5×1019 (d) 2.4×1033 (e) 8.1×1024

Ans: (b) In 1 s, the energy emitted by the laser is E = 10-2 J/s × 1 s = 10-2 J. The energy of one photon of frequency f = c/λ is hf. In this case λ = 540 nm, so the number of photons is

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