Partial differentiation

Partial differentiation

Wave equation 1 = Example: Show that the following functions are solutions of the wave equation.

In fact, we can show that any functions with the form, for any differentiable functions f(u) and g(u) are solutions of the wave equation.

y 15.1. IDENTIFY: vy =. v = f λ = λ/ T. t π π v π SET UP: Acos ( x vt) = + A sin ( x vt) t λ λ λ EXECUTE: (a) x t π λ π λ A cos π = + A cos x t = + A cos ( x vt) where = λ f = v has been used. λ T λ T λ T y π v π (b) vy = = Asin ( x vt). t λ λ (c) The speed is the greatest when the sine is 1, and that speed is πva / λ. This will be equal to v if A = λ/ π, less than v if A < λ/π and greater than v if A > λ/ π. EVALUATE: The propagation speed applies to all points on the string. The transverse speed of a particle of the string depends on both x and t.

E x a m p le 3.. (1) Find the amplitude, frequency, wavelength, and sp eed of propagation of t he wave describ ed by t he equation y = 0. cos [π(5t x)]. Here, the units of lengt h and t ime are t aken t o be meter and second, resp ect ively. () W hen a sinusoidal wave of amplitude 0.1 m and frequency Hz t ravels at a sp eed of m/ s in t he x direct ion, derive t he expression for t he displacem ent y at p osit ion x at t im e t by using an integer, n. Here, we assume that the displacement at t he origin (x = 0) at t ime t = 0 is zero, i.e., y = 0.

Solu t ion (1) From Eq. (3.10), the amplitude is A = 0. m, t he period is T = 0.4 s, and t he wavelengt h is λ = 1.0 m. From these values, the frequency, f = 1/ T =.5 Hz, and the propagat ion speed of the wave, V = f λ =.5 m/ s. () Using A = 0.1 m, T = 0.5 s and λ = VT = 1 m (b ecause V = m/ s), we have y(x, t) = 0.1 cos 4π t + x + n + 1 π = 0.1 sin 4π t + x + nπ 4.

15.3. IDENTIFY: The average power carried by the wave depends on the mass density of the wire and the tension in it, as well as on the square of both the frequency and amplitude of the wave (the target variable). SET UP: EXECUTE: Solving P 1 av =, µ Fω A v = F µ. 1/ P 1 av = F A for A gives P A av = µ ω ω µ F ω = π f = π (69. 0 Hz) = 433. 5 rad/s. The tension is F = 94. 0 N and F 94.0 N 4 µ = = = 3.883 10 kg/m. v (49 m/s) 1/. P av = 0. 365 W. v = F so µ (0.365 W) 3 A = = 4. 51 10 m = 4.51 mm 4 (433.5 rad/s) (3.883 10 kg/m)(94.0 N) EVALUATE: Vibrations of strings and wires normally have small amplitudes, which this wave does. 15.4. IDENTIFY: The average power (the target variable) is proportional to the square of the frequency of the wave and therefore it is inversely proportional to the square of the wavelength. SET UP: EXECUTE: to 1. λ P 1 av = µ Fω A where ω = π f. The wave speed is v = F. µ v π F ω = π f = π = so λ λ µ Therefore av,1λ 1 = av,λ P P and 1 4π F Pav = µ F. A This shows that P av is proportional λ µ 1 λ1 λ Pav, = P av,1 = (0. 400 W) = 0. 100 W. λ λ1 EVALUATE: The wavelength is increased by a factor of, so the power is decreased by a factor of = 4.

15.3. IDENTIFY: Apply the principle of superposition. SET UP: The net displacement is the algebraic sum of the displacements due to each pulse. EXECUTE: The shape of the string at each specified time is shown in Figure 15.3. EVALUATE: The pulses interfere when they overlap but resume their original shape after they have completely passed through each other. Figure 15.3

15.38. IDENTIFY: Evaluate SET UP: sin kx = k cos kx. x EXECUTE: (a) y/ x and y/ t and see if Eq. (15.1) is satisfied for v = ω/ k. cos kx = k sin kx. x y = k [ A sw sin ωt]sin kx, x, t sinωt = ω cos ωt. t y = ω [ A sin ωt]sin kx, sw cosωt = ω sinωt t so for y( x, t ) to be a solution ω ω of Eq. (15.1), k = and v =. v k (b) A standing wave is built up by the superposition of traveling waves, to which the relationship v = λ/ k applies. EVALUATE: y( x, t) = ( A sin kx)sinωt is a solution of the wave equation because it is a sum of SW solutions to the wave equation. 15.39. IDENTIFY: Evaluate y/ x and y/ t and show that Eq. (15.1) is satisfied. y1 y SET UP: ( y1 + y) = + x x x and y1 y ( y1 + y) = + t t t y y1 y = + y y1 y EXECUTE: and = + x x x t t t. The functions y 1 and y are given as being solutions to the wave equation, so y y1 y 1 y1 1 y 1 y 1 y 1 y = + = + = + = and so y = y 1 + y is a x x x v t v t v t t v t solution of Eq. (15.1). EVALUATE: The wave equation is a linear equation, as it is linear in the derivatives, and differentiation is a linear operation.

15.59. IDENTIFY: The frequency of the fundamental (the target variable) depends on the tension in the wire. The bar is in rotational equilibrium so the torques on it must balance. SET UP: v = F and f = v. Σ τ z = 0. µ λ EXECUTE: λ = L = 0. 660 m. The tension F in the wire is found by applying the rotational equilibrium methods of Chapter 11. Let l be the length of the bar. Then Σ τ z = 0 with the axis at the hinge gives 1 Fl cos30 = lmg sin 30. F 17. 3 N v = = = 1. 37 m/s. µ (0. 090 kg/0. 330 m) mg tan 30 (45.0 kg)(9.80 m/s ) tan 30 F = = = 17.3 N. EVALUATE: This is an audible frequency for humans. v 1. 37 m/s f = = = 3. 4 Hz λ 0. 660 m

This is an audible frequency for humans. 15.60. IDENTIFY: The mass of the planet (the target variable) determines g at its surface, which in turn determines the weight of the lead object hanging from the string. The weight is the tension in the string, which determines the speed of a wave pulse on that string. m SET UP: At the surface of the planet EXECUTE: On earth, v = Mg µ p p g = G. The pulse speed is v = F. R µ 4.00 m 1.056 10 m/s. v = = 0.0390 s and the mass of the lead weight is 0.080 kg 3 3 µ = = 7.00 10 kg/m. F = Mg, so 4.00 m

15.74. IDENTIFY: The displacement of the string at any point is y( x, t) = ( ASW sin kx)sin ωt. For the fundamental mode λ = L, so at the midpoint of the string sin kx = sin( π / λ)( L /) = 1, and y = A SW sin ωt. The transverse velocity is v = y/ t and the transverse acceleration is a = v / t. y y SET UP: Taking derivatives gives vy = = ω ASW cos ωt t, with maximum value vy, max = ω ASW, and vy ay = = ω A SW sin ωt, with maximum value a t y, max = ω ASW. EXECUTE: SW y, max y, max y, max 3 3 ω = a / v = (8. 40 10 m/s )/(3. 80 m/s) =. 1 10 rad/s, and then 3 3 A = v / ω = (3.80 m/s)/(.1 10 rad/s) = 1.7 10 m. (b) v = λ f = ( L)( ω/ π ) = Lω/ π = (0. 386 m)(. 1 10 rad/s) / π = 7 m/s. EVALUATE: The maximum transverse velocity and acceleration will have different (smaller) values at other points on the string. 3 y y

16.39. IDENTIFY: The beat is due to a difference in the frequencies of the two sounds. SET UP: f beat = f 1 f. Tightening the string increases the wave speed for transverse waves on the string and this in turn increases the frequency. EXECUTE: (a) If the beat frequency increases when she raises her frequency by tightening the string, it must be that her frequency is 433 Hz, 3 Hz above concert A. (b) She needs to lower her frequency by loosening her string. EVALUATE: The beat would only be audible if the two sounds are quite close in frequency. A musician with a good sense of pitch can come very close to the correct frequency just from hearing the tone.

v + vl 16.45. IDENTIFY: Apply the Doppler shift equation fl = fs. v + vs SET UP: The positive direction is from listener to source. f S = 39 Hz. v + vl 344 m/s 15. 0 m/s (a) v S = 0. v L = 15. 0 m/s. fl = fs = (39 Hz) = 375 Hz v + vs 344 m/s v + vl 344 m/s + 15. 0 m/s (b) v S = + 35. 0 m/s. v L = + 15. 0 m/s. fl = fs = (39 Hz) = 371 Hz v + vs 344 m/s + 35. 0 m/s (c) fbeat = f1 f = 4 Hz EVALUATE: The distance between whistle A and the listener is increasing, and for whistle A f L < f S. The distance between whistle B and the listener is also increasing, and for whistle B f L < f S.

v 16.64. IDENTIFY: The harmonics of the string are fn = nf1 = n, where l is the length of the string. The tube l is a stopped pipe and its standing wave frequencies are given by Eq. (16.). For the string, v = F / µ, where F is the tension in the string. SET UP: The length of the string is d = L/10, so its third harmonic has frequency pipe v The stopped pipe has length L, so its first harmonic has frequency f s 1 =. 4L string pipe 1 EXECUTE: (a) Equating f 1 and f 1 and using d = L/10 gives F = µ vs. 3600 f string 3 1 = 3 F/ µ. d (b) If the tension is doubled, all the frequencies of the string will increase by a factor of. In particular, the third harmonic of the string will no longer be in resonance with the first harmonic of the pipe because the frequencies will no longer match, so the sound produced by the instrument will be diminished. (c) The string will be in resonance with a standing wave in the pipe when their frequencies are equal. Using pipe string 1 f1 pipe f = 3, the frequencies of the pipe are nf1 = 3nf1 (where n = 1, 3, 5, ). Setting this string equal to the frequencies of the string n f 1, the harmonics of the string are n = 3n = 3, 9,15, The nth harmonic of the pipe is in resonance with the 3nth harmonic of the string. EVALUATE: Each standing wave for the air column is in resonance with a standing wave on the string. But the reverse is not true; not all standing waves of the string are in resonance with a harmonic of the pipe. string

same in both cases. 16.8. IDENTIFY and SET UP: Use Figure (16.37) to relate α and T. Use this in Eq. (16.31) to eliminate sinα. EXECUTE: Eq. (16.31): α = v vs From Figure 16.37 S sin / h Combining these equations we get = v T v T S v 1 ( v/ v ) = and v = h 1 v T / h hv vs = as was to be shown. h v T S S v/ v S 1 ( v/ v ) S and tan α = h/ v T. And h = T v 1 ( v/ v ) S sinα sinα tan α = =. cosα 1 sin α EVALUATE: For a given h, the faster the speed v S of the plane, the greater is the delay time T. The maximum delay time is h/ v, and T approaches this value as vs. T 0 as v v S..

E x a m p le 3.5. Let us discuss a sinusoidal wave moving in t he + x direct ion. y 1 (x, t) = A sin π T t x v. A wall locat ed at x = L refl ect s t his wave. Answer t he following quest ions for each of t he following two cases: (a) T he wall is a fi xed end. (b) T he wall is a free end. (1) F ind t he expression for t he refl ect ed wave. () F ind t he expression for t he result ant wave produced by t he incident wave and t he refl ect ed wave. Not e t hat a fi xed end is an end where t he am plit u de of t he result ant wave of t he incident and refl ect ed waves vanishes at all tim es, whereas a free end is an end where the displacem ent of the refl ect ed wave is equal t o t hat of t he incident wave.

Solu t ion (1) T he refl ect ed wave moves in t he x direct ion, and has t he same amplit ude, p eriod, and velocity as t he incident wave. T herefore, the refl ected wave can be written as y (x, t) = A sin π T t + x v + β, whereβis a const ant t hat is t o be det ermined by the condit ion that the wave should satisfy at the end x = L (a b oundary condit ion). (a) In t he case where x = L is a fi xed end, t he displacement y must always sat isfy t he condit ion y(l, t) = y 1 (L, t) + y (L, t) = 0. T herefore, sin π T t L v + sin π T t + L v + β = 0. From t his equat ion, β is det ermined t o beβ= T L v, and we obt ain y (x, t) = A sin = A sin π T π T t + x L v t + x L v + π. (b) On t he ot her hand, in t he case where x = L is a free end, t he displacement y must sat isfy t he condit ion. y 1 (L, t) = y (L, t). T herefore, sin π T t L v = sin π T t + L v + β. From t his equat ion,βis det ermined to beβ= L v, and we obt ain y (x, t) = A sin π T t + x L v () T he resultant wave is y(x, t) = y 1 (x, t) + y (x, t). T hen, we have (a) y(x, t) = A sin (b) y(x, t) = A cos π T L x v π T L x v cos sin π T π T. t L v t L v We can see t hat t hese result ant waves are st anding waves, oscillat ing independent ly in t ime and in space..