PHYS1169: Tutorial 8 Solutions

Transcription

1 PHY69: Tutorial 8 olutions Wae Motion ) Let us consier a point P on the wae with a phase φ, so y cosϕ cos( x ± ωt) At t0, this point has position x0, so ϕ x0 ± ωt0 Now, at some later time t, the position x of this point on the wae P can be foun byϕ x ωt, since the point shoul hae the same isplacement y an same phase φ Thus, we hae x ( ϕ m ωt) an x0 ( ϕ m ωt0 ) The elocity of the wae is the istance traelle oer the time taen: x x x0 t t t With the aboe expressions for x an 0 x 0, this gies [( ϕ m ωt) ( ϕ m ωt0 )] ω m t t0 But, expressing the angular frequency ω in terms of the frequency f, ω f an the wae number in terms of the waelength λ,, gies m fλ Thus, a point λ with a gien phase will moe in the positie x-irection with elocity fλ (corresponing to the lower sign in the aboe analysis), while the elocity is fλ, with trael in the negatie x-irection if the positie sign is use F ) Use, where is the elocity of the wae, F is the tension in the wire, an is the mass per unit length, to fin the elocity of the wae in each wire For steel, For copper, m ρ V ρ πr L ρ πr ρ πr L L L

2 π ( 05 0 ) π ( 05 0 ) 67 0 gm F 56ms gm F 6ms The time for the wae to trael the 30m of steel wire is thus t 0 9s, 56 0 an similarly for the copper wire t 0 37s The total time to trael 6 along the two wires is t t + t 0 39s 3) At x 0, y ( 5cm) cos( 503t) (5cm) cos(503t), since cos x is an een function (ie cos( x) cos x ) (Not to scale) The perio of the wae is T 05s This agrees with T 0 5s ω 503 ) a) tart with y Asin( x ωt) an use ω to get y Asin ( x t) b) Then use in a) to get y Asin λ ( x t) λ x c) Then use fλ in b) to gie y Asin ( ft) λ x ) Finally, use λ in c) to get y Asin f f ( t) 5) The wae y ( 05m)sin(03x 0t) is in the form y Asin( x ωt), so by inspection we can etermine that: a) amplitue A 05m b) angular frequency ω 0s c) wae number 03m

3 ) waelength λ 09m 03 ω e) wae spee 0 333ms 03 f) from question, the negatie sign in the phase relationship x ωt implies that the wae is traelling in the positie x-irection 6) The wae y ( 008m)sin(0x 30t) is in the form y Asin( x ωt), so by inspection 0m an ω 30s ω a) Velocity of the wae is 30 5ms 0 b) The transerse elocity is the elocity perpenicular to the irection of propagation of the wae, that is, the elocity in the y-irection The elocity in the y-irection is just the change in y position with respect to time, for a gien x point in the waes y path, t (that is, the eriatie of y with respect to t, treating x as a t x, cons tan t constant to consier the transerse isplacement at a fixe x point in the wae s path) Thus, y t 008 ( 0) cos(0x 30t), t x, cons tan t using the stanar result ( Asin( at b) ) Aa cos( at b) t The transerse elocity as a function of time is then t ( ms )cos(0x 30t) The expression will hae its maximum elocity when the cosine factor is -, so the maximum transerse elocity reache is t ms max oun Waes ) Let the source of the lightening strie be a istance away from the obserer The time taen for the soun at s 33ms to reach the obserer is t s imilarly, s 8 the time taen for the light to reach the obserer at c 30 0 ms is t L The c time ifference between the light an the soun arrial times is then t t s t L s c Because the light traels much, much faster than the soun, the time for the light to reach the obserer t L is so much smaller than the time for the soun to reach the obserer that we can neglect it (if your unsure, plug in the numbers an see!) t s t 6 s so that s t m Thus, the source of the lightning is 556m away s

4 B ) Use, where is the elocity of the soun wae, B is the bul moulus of ρ the meium, an ρ is the ensity of the meium ms ) The intensity soun leel of soun wae is expresse relatie to the threshol leel of human hearing, which is taen as I 0 0 Wm The soun leel in ecibels is 6 I x0 L( B) 0log0 0log0 66B I 0 0 ) Firstly, fin the intensity in Wm for the soun leel of 0B, ia I I L ( B) 0B 0log 0 0log0 I 0 0 I I 0 Wm 0 This intensity can then be relate to the pressure amplitue p of the soun wae ia the relation I p ρ, where ρ is the ensity of the meium an is the elocity of the wae Then, p I ρ Pa Extra Problems E) a) At x 5cm an t s, the phase of the first wae an secon wae are: φ 0x 30t ra φ 5x 0t ra Thus, there is a phase ifference φ φ φ 5ra b) At t s, the aition of the two waes is y T y + y sin(0x 60) + sin(5x 80) We can use the trig relation to express this as the prouct of a sine an a cosine factor, A + B A B sin A + sin B sin cos that gies y T sin( 5x 70) cos( 5x 0) ) Thus y T will be zero wheneer the sine an the cosine terms are zero We see the smallest positie x such that y T is zero, so we may tae the smallest positie x so that the sine term is zero an the smallest positie x that the cosine term is zero, then the smaller of the two will be our esire alue of x Now, sinθ is zero for any multiple of π, so we can fin the integer so that 5x 70 π has the smallest positie x alue By trial an error, this is -, 70 π 0885 for which x cm imilarly, cosθ is zero for any o 5 5

5 multiple of π/, so we can fin the alue of m that gie the smallest positie x with ( m + ) π 5x 0 This happens when m -3 an then x 0858cm Thus the first positie alue of x at which the two waes a to zero near the origin is 00393cm T E) Use with elocity 50 m/s an mass per unit length 006g 00gm The tension in the string is then 5m T N is tan ce 5m E3) The elocity of the waes are 05ms The time 0s 0ib frequency of the wae is the number of ibrations per secon, f Hz 3 30s Then the waelength is simply λ 039m f E) a) We can use the general form for a wae traelling in the negatie x-irection, y Asin( x + ωt φ) where the amplitue A 008m, wae number 785m, an angular λ 08 frequency ω f 3 6π s Thus, y ( 008m)sin(785x + 6πt φ) We can fin the phase ф by y ( x 0, t 0) 0, so 0 008sin( φ ) which allows us to tae ф 0 The expression for the wae is then y ( 008m)sin(785x + 6πt) b) As aboe, except now y ( x 0, t 0) 0, so 0 008sin(785 φ) an we may tae φ 7 85 so the wae is escribe as y ( 008m)sin(785x + 6π t 785) E5) a) The waelength is simply λ m 0 8m, so the wae number can be foun 5 from 785m The angular frequency is just λ 08 ω s The wae is then escribe using y Asin( x ωt) to gie y ( 00m)sin(785x 005t) 8 b) The frequency of the wae is f 60Hz Expressing the wae explicitly λ 08 showing the waelength an the frequency is x x y Asin ft (00m)sin 60t λ 08 E6) The pressure amplitue is relate to the isplacement amplitue s ia p ρ m s m p m m

6 33 The elocity of the wae is 33 m/s, the wae length λ 0033m 3 f 0 0, so the wae number is 83m, an the ensity is ρ gm λ 0033 Thus, the isplacement amplitue is pm 0 0 sm 55 0 m ρ E7) The pressure ariation of the wae is just a sinusoial wae as p pm sin x ωt The wae number is 68m, an the angular frequency is λ 0 ω s Therefore, p ( 0Pa)sin 68x 6 0 t