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# 7soln7 - Physics 315 Oscillations and Waves Homework 7...

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Unformatted text preview: Physics 315: Oscillations and Waves Homework 7: Solutions 1. Consider ¯ F (k ) = = 1 2π 1 2π x2 exp − 2 2 σx −∞ ∞ ∞ exp(i k x) dx (1) x2 exp − + i k x dx. 2 2 σx −∞ Now, completing the square, 2 4 x2 −(x − i σx k )2 − σx k 2 − + ikx = , 2 2 2 σx 2 σx 2 (x − i σx k )2 exp − dx, 2 2 σx −∞ √ 2 where σk = 1/σx . Let z = (x − i σx k )/ 2 σx . It follows that √ k2 2 σx ∞ −z2 ¯ (k ) = exp − F e dz. 2 2 σk 2π −∞ 2 ¯ (k ) = exp − k F 2 2 σk (2) so 1 2π ∞ (3) (4) However, 2 ∞ e−z −∞ dz = √ π , so 1 k2 exp − 2 2 2 σk 2π σk . (5) ¯ F (k ) = Now, from de Moirve’s theorem, 1 ¯ F (k ) = 2π ∞ −∞ exp − x2 2 2 σx [cos(k x) + i sin(k x)] dx, (6) Hence, ¯ Re[F (k )] ¯ Im[F (k )] It thus follows that C (k ) = S (k ) 2. Consider I= −∞ = = 1 2π 1 2π ∞ −∞ ∞ exp − x2 2 2 σx cos(k x) dx = C (k ), sin(k x) dx = S (k ). (7) (8) x2 exp − 2 2 σx −∞ ¯ Re[F (k )] = ¯ Im[F (k )] = 0. ∞ k2 1 exp − 2 2 2 σk 2π σk , (9) (10) = 1 k2 exp − 2 2 2 σk 2π σk dk. (11) √ Let z = k/ 2 σk . It follows that √ √ ∞ 2 2 σk 2 σk √ I= e−z dz = π = 1. 2 2 2π σk −∞ 2π σk 3. The cosine Fourier transform of the function is 1 C (ω ) = 2π 1 F (t) cos(ω t) dt = 2π −∞ ∞ ∞ 0 (12) exp − t 2τ cos(ω t) dt. (13) Likewise, the sine Fourier transform of the function is 1 S (ω ) = 2π Let us deﬁne ¯ F (ω ) = C (ω ) + i S (ω ) = = 1 2π 1 2π ∞ 0 ∞ 0 1 F (t) sin(ω t) dt = 2π −∞ ∞ ∞ 0 exp − t 2τ sin(ω t) dt. (14) exp − exp − t 2τ exp (i ω t) dt dt (15) t + iωt 2τ where use has been made of de Moirve’s theorem. Thus, e−t/(2 τ )+i ω t 1 ¯ F (ω ) = 2π −1/(2 τ ) + i ω which yields C (ω ) S (ω ) 4. Let I= −∞ ∞ = 0 1 τ τ 1 + i2ωτ = , π 1 − i2ωτ π 1 + (2 ω τ )2 τ 1 , π 1 + (2 ω τ )2 2ωτ τ . π 1 + (2 ω τ )2 (16) ¯ = Re[F (ω )] = ¯ = Im[F (ω )] = (17) (18) ∞ F 2 (t) dt. (19) It follows that ∞ ∞ ∞ 2 I= −∞ −∞ C (ω ) cos(ω t) dω + −∞ S (ω ) sin(ω t) dω dt. (20) This can also be written ∞ ∞ ∞ I = −∞ ∞ −∞ C (ω ) cos(ω t) dω −∞ ∞ C (ω ′ ) cos(ω ′ t) dω ′ dt ∞ +2 −∞ ∞ −∞ ∞ C (ω ) cos(ω t) dω −∞ ∞ S (ω ′ ) sin(ω ′ t) dω ′ dt S (ω ′ ) sin(ω ′ t) dω ′ dt. (21) −∞ + −∞ −∞ S (ω ) sin(ω t) dω Rearranging the order of integration, we obtain ∞ ∞ ∞ I = −∞ C (ω ) C (ω ) −∞ ∞ ∞ ′ −∞ ∞ −∞ ′ cos(ω t) cos(ω ′ t) dt dω dω ′ ∞ +2 −∞ ∞ C (ω ) S (ω ) −∞ ∞ cos(ω t) sin(ω ′ t) dt dω dω ′ sin(ω t) sin(ω ′ t) dt dω dω ′ . (22) −∞ + −∞ −∞ S (ω ) S (ω ′ ) Making use of Eqs. (8.29)–(8.31) in the lecture notes, we obtain ∞ ∞ I =π +π −∞ −∞ −∞ ∞ ∞ C (ω ) C (ω ′ ) [δ (ω − ω ′ ) + δ (ω + ω ′ )] dω dω ′ S (ω ) S (ω ′ ) [δ (ω − ω ′ ) − δ (ω + ω ′ )] dω dω ′ . (23) −∞ Making use of Eq. (8.26) in the lecture notes, we obtain ∞ ∞ I = π +π −∞ −∞ −∞ ∞ ∞ [C (ω ) C (ω ) + C (ω ) C (−ω )] dω [S (ω ) S (ω ) − S (ω ) S (−ω )] dω. (24) −∞ But, C (−ω ) = C (ω ) and S (−ω ) = −S (ω ), so ∞ I = 2π −∞ C 2 (ω ) + S 2 (ω ) dω. (25) 5. If ∞ F (t) = −∞ ∞ ¯ F (ω ) cos(ω t) dω, ¯ G(ω ) cos(ω t) dω, −∞ ∞ (26) (27) (28) G(t) H (t) then ¯ F (ω ) ¯ G(ω ) ¯ H (ω ) = = −∞ ¯ H (ω ) cos(ω t) dω, = = = 1 2π 1 2π 1 2π ∞ F (t) cos(ω t) dt, −∞ ∞ (29) (30) (31) G(t) cos(ω t) dt, −∞ ∞ H (t) cos(ω t) dt. −∞ Let H (t) = F (t) G(t). It follows that ¯ H (ω ) = = 1 2π 1 2π ∞ F (t) G(t) cos(ω t) dω −∞ ∞ ∞ ∞ ∞ (32) ¯ G(ω ′′ ) cos(ω ′′ t) dω ′′ cos(ω t) dt. −∞ ¯ F (ω ′ ) cos(ω ′ t) dω ′ −∞ Rearranging the order of integration, we obtain ∞ ¯ H (ω ) = −∞ 1 ¯ ¯ F (ω ′ ) G(ω ′′ ) 2π −∞ ∞ ∞ cos(ω t) cos(ω ′ t) cos(ω ′′ t) dt dω ′ dω ′′ . −∞ (33) Now, from (8.29) in the lecture notes, 1 2π ∞ cos(ω t) cos(ω t) cos(ω t) dt −∞ ′ ′′ = 1 4π + 1 4π ∞ cos[(ω + ω ′ ) t] cos(ω ′′ t) dt −∞ ∞ −∞ cos[(ω − ω ′ ) t] cos(ω ′′ t) dt = 1 1 δ (ω + ω ′ − ω ′′ ) + δ (ω + ω ′ + ω ′′ ) 4 4 1 1 + δ (ω − ω ′ − ω ′′ ) + δ (ω − ω ′ + ω ′′ ). 4 4 (34) Thus, making use of Eq. (8.26) in the lecture notes, we obtain ¯ H (ω ) = 1 4 1 + 4 1 ¯ ¯ F (ω ′ ) G(ω + ω ′ ) dω ′ + 4 −∞ 1 ¯ ¯ F (ω ′ ) G(ω − ω ′ ) dω ′ + 4 −∞ ∞ ∞ ∞ −∞ ∞ ¯ ¯ F (ω ′ ) G(−ω − ω ′ ) dω ′ ¯ ¯ F (ω ′ ) G(−ω + ω ′ ) dω ′ . (35) −∞ ¯ ¯ However, since G(−ω ) = G(ω ), we can also write 1 ¯ H (ω ) = 2 ∞ −∞ ¯ ¯ ¯ F (ω ′ ) G(ω ′ + ω ) + G(ω ′ − ω ) dω ′ . (36) Now, if F (t) = cos(ω0 t) then 1 ¯ F (ω ) = 2π So ¯ H (ω ) = 1 4 + which yields 1¯ 1¯ ¯ ¯ ¯ G(ω0 + ω ) + G(ω0 − ω ) + G(−ω0 + ω ) + G(−ω0 − ω ) . H (ω ) = 4 4 (39) ¯ (−ω ) = G(ω ), so we get ¯ But, G 1¯ ¯ ¯ H (ω ) = G(ω − ω0 ) + G(ω + ω0 ) . 2 (40) 1 4 ∞ −∞ ∞ ∞ cos(ω0 t) cos(ω t) dt = −∞ 1 [δ (ω0 − ω ) + δ (ω0 + ω )] . 2 (37) ¯ ¯ δ (ω0 − ω ′ ) G(ω ′ + ω ) + G(ω ′ − ω ) dω ′ ¯ ¯ δ (ω0 + ω ′ ) G(ω ′ + ω ) + G(ω ′ − ω ) dω ′ , (38) −∞ ...
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