# pset1sol - 22.02 Problem Set 1 Solution 1 Wave Interference...

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Unformatted text preview: 22.02 Problem Set 1 Solution 1. Wave Interference Solution S 2 S 1 D b Assume S 1 and S 2 are two identical sources. Their electric field components at point P can be expressed as: E 1 ( t ) = E cos ω t (1) t E 2 ( t ) = E cos( ω + φ ) (2) E 1 E 2 E p φ Use phasor representation, the resulting interference wave electric field component is: E p 2 = E 2 + E 2 − 2 E 2 cos( π − φ ) E p 2 = 2 E 2 + 2 E 2 cos φ = 2 E 2 (1 + cos φ ) E p 2 = 4 E 2 cos 2 φ (3) 2 The phase difference φ is associated with path difference as: φ r 2 − r 1 = 2 π λ d sin θ If D >> d , r 2 − r 1 ≈ d sin θ ⇒ φ ≈ 2 π λ (4) 22.02 Problem Set 1 Solution E 2 Also we know the intensity of the wave is I = 2 m C E 2 Thus I = p p 2 µ C E 2 Insert (3), we get I p = 2 µ C × 4cos 2 φ 2 = 4 I 0 cos 2 φ 2 (5) In summary, I p = 4 I 0 cos 2 φ , φ ≈ 2 π d sin θ 2 λ 2. Solution (a) Transverse Wave: displacement of medium is perpendicular to the direction of travel of the wave. The displacement of a particle on the string at location x and time t can be expressed as: y ( x , t ) = A sin 2 π ( t / T + x / λ ) (1) 2 π Among them, A: amplitude, T: period, λ : wavelength, k = λ : wave number, 1 λ f = T : frequency, v = T : velocity. Compare (1) with y = 0.3sin [π (0.5 x − 50 t ) ] = 0.3sin [ 2 π (0.25 x − 25 t ) ] , we get π A 0 3 T 0 04 / λ 4 k 1/ f 25(1/ ) = . cm , = . cm s , = cm , = 4 cm v = 0.04 cm / s = 100 cm / s (b) The transverse speed of a particle at x is v ( x , t ) = dy ( x , t ) = 15 π cos[ π (0.5 x − 50 t )] dt The maximum speed is 15 π cm / s ....
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pset1sol - 22.02 Problem Set 1 Solution 1 Wave Interference...

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