School of Mathematics & Statistics MT 4507/ Classical Mechanics - Solutions Sheet 1

Transcription

1 T. Neukirch Sem /3 School of Mathematics & Statistics MT 457/587 - Classical Mechanics - Solutions Sheet. (i) A = (,, ), B = (4,, 3) C = (,, ) (a) The magnitudes of A, B and C are A A = A = + + = 9 = A = 3 B B = B = = 5 = B = 5 C C = C = + + = 9 = C = 3 (b) A B = AB cos θ, A = A, B = B, θ = angle between A and B. (,, ) (4,, 3) = = = 3 5 cos θ so cos θ = /3 = θ = 48. (c) B C = B C sin θ, θ = angle between B and C. Now and so B C = (4,, 3) (,, ) = ( 3, 4, 4) B C = = = 5 3 sin θ = θ = 8.3 (a) D E = D (αd) = αd D =, since D is parallel to D. (b) F = (F x, F y, ), G = (G x, G y, ) F G = (F y G y, G x F x, F x G y F y G x ) = (,, F x G y F y G x ), so F G is in the ẑ direction. Alternatively use the right-hand rule: align your index finger with the first vector (F), your second finger with the second vector (G) then your thumb will define the direction F G. (c) Write D = D + D where D and D are the components of D perpendicular and parallel to H, respectively. D H = (D + D ) H = D H + D H }{{} =

2 The magnitude of D H is since the angle between D and H is π/. D H = D H = D H sin(π/) = D H Alternatively, we can resolve H into components H and H, which are parallel or perpendicular to D, respectively. Then and D H = D (H + H ) = D H +D H }{{} = D H = D H = DH sin(π/) = DH.. Use the right-hand rule, or otherwise (a) ˆx ŷ = ẑ (b) ŷ (ˆx ẑ) = ŷ ( ŷ) =, since ŷ and ŷ are parallel/antiparallel. (c) ˆx (ˆx ŷ) = ˆx ẑ = ŷ. (d) (ˆx ˆx) ŷ = ŷ =, since ˆx and ˆx are parallel. (e) ˆr ˆθ = ẑ (f) ẑ (ẑ ˆr) = ẑ ˆθ = ˆr 3. (a) (i) f(x) = x n ; df = nxn f(x) = sin(ax) + cos(bx); df = a cos(ax) b sin(bx) f(x) = x n sin(x); df = nxn sin(x) + x n cos(x) (iv) g(x) = exp[f(x)], f(x) = x n sin(x) dg = dg df df = (nxn sin(x) + x n cos(x)) exp[x n sin(x)] (v) f(x) = cos(ax ); df = sin(ax )(ax) = ax sin(ax ) (b) For any function φ(x) we have so d φ dx = so dφ dx = f(x)dφ. d ( ) dφ = f(x) d ( f(x) dφ ) [ ] df dφ = f(x) dx dx + φ f(x)d 4. φ is given by φ = sin(kx) + x 3 y + ye z (i) d dx f (x) d df + f(x) A = φ = ( φ x, φ y, φ ) z d A = (k cos(kx) + 3x y, x 3 + e z, ye z ) = f (x) d φ df dφ +f(x)

3 A = A x x + A y y + A z z = ( k sin(kx) + 6xy) + () + (ye z ) = k sin(kx) + 6xy + ye z A = φ =, this is a vector identity. We can confirm this by explicit evaluation : ( A) x = A z y A y z = ez e z = ( A) y = A x z A z x = = ( A) z = A y x A x y = 3x 3x = so A =. φ = ( φ) = A = k sin(kx) + 6xy + ye z from above. ψ = (x ln(z) + cos(x) cos(y/), y ln(z) sin(x) sin(y/) +, (x + y )/z) Now ψ z = x + y z = ψ = (x + y ) ln(z) + f(x, y) Now take / x : ψ = x ln(z) + x x Equate this with the ˆx component of ψ to give x ln(z) + x = x ln(z) + cos(x) cos(y/) = x = cos(x) cos(y/) i.e. f(x, y) = sin(x) cos(y/) + g(y) = ψ = (x + y ) ln(z) + sin(x) cos(y/) + g(y) Take / y, and equate with ŷ component of ψ : = dg ψ y =, = g = y + c, c = constant = y ln(z) sin(x) sin(y/) + dg = y ln(z) sin(x) sin(y/) + So, finally ψ = (x + y ) ln(z) + sin(x) cos(y/) + y + c 5. (i) A = (, x, y ) Calculate C A dl, where C composed of straight line segments C from (,, ) to (,, ) and C from (,, ) to (,, ).

4 Notice that A dl = C A dl + C A dl C On C, dl = ˆx and y = z = C A dl = A x (x,, ) = = [ x ] C is along the line y = z, x =, so dz =. On C we have dl = ŷ + ẑdz = ŷ + ẑ dz = (ŷ + ẑ) and we have parametrized the integral in terms of y, y. So C A dl =,x= (A y + A z ) = Combining the two integrals we obtain,x= x + C A dl = 4 3,x= = y = [ y ] + [ y 3 /3 ] = 3 A = (x, yz, x sin y + z), R is the rectangle (, ), (, ), (, π), (, π), in the plane z =. R A ˆndS = Since z = on R we have A z = x sin(y) +, so R R A ẑds = R A z π π A z = x sin(y) = x sin(y) y= x= = [ x / ] [ ] ( ) π cos(y) = ( ( )) = Ψ = r cos θ, S ΨdV, S = sphere r a In spherical coordinates (r, θ, φ), we have dv = r sin θdrdθdφ, so S ΨdV = = π φ= a π a θ= r= π r 4 dr r cos θ r sin θdrdθdφ cos θ sin θdθ = [ r 5 /5 ] a[ cos 3 θ/3 ] π[ ] π φ π dφ = a5 5 π 3 = 4π 5 a5

5 6. (i) d x + ω x =, ; x = A sin(ωt) + B cos(ωt), A, B constants. x = at t = = B =, x = A sin(ωt) So = at t = : = Aω cos(ωt) = A = /ω x = ω sin(ωt) d x ω x = = x = A exp(ωt) + B exp( ωt) x = at t = = A + B = as t : =, x = ln(t) + c, c =constant. t x = at t = = c = (iv) + ωy = and ωx = so x = A sin(ωt) + B cos(ωt). = ωa exp(ωt) ωb exp( ωt) = A = and B = from above. = d x x = at t = : = B =, A undetermined. x = A sin(ωt) + cos(ωt) y = at t = : y = ω x = exp( ωt) x = ln(t) + = ω = ω x (from rearranging the first equation) so y = (Aω cos(ωt) ω sin(ωt)) = A cos(ωt) + sin(ωt) = A = ω x = sin(ωt) + cos(ωt), y = cos(ωt) + sin(ωt) 7. (i) f(x) = ln( + x), about x = ; f(x) f() + x df x= +...; df = df = + x x= =

6 f(x) ln() + x = x f(x) = sin(x), about x = π/4, write x = π/4 + δx. Now df f(π/4 + δx) = f(π/4) +δx df }{{} x=π/ =/ = cos(x), df x=π/4 = cos(π/4) = /. f(x) = + δx = + (x π/4) f(x, y) = exp(x + y), about (x, y) = (, ). f( + δx, + δy) = f(, ) + δx x (,) + δy y (,) +... x = 3x exp(x + y), x (,) = e y = exp(x + y), y (,) = e (iv) = f( + δx, δy) = e + eδx + eδy = e + e(x ) + ey Expand d f/ + sin(f) = for small f and solve. Write f(x) = εf (x) + ε f (x) +..., where ε and f n O(). Subsitute in the equation to get ε d f + d f ε sin(εf + ε f +...) = }{{} εf +O(ε ) Collecting the leading terms (of order ε) we get d f + f = (like 6(i) with ω =, x f, t x) f = A sin(x) + B sin(x) and f εf + O(ε ) = εa sin(x) + εb cos(x) and df = εa cos(x) εb sin(x) Since df = at x = = A =, so because f = δ( ) at x =, so εb = δ. f = εb cos(x) = δ cos(x)

7 8. A = B = 4 (i) C = A B = If λ is an eigenvalue of A and X is an eigenvector, then A X = λx = λi X, I = 4 = 8 4 = (A λi) X = This has non-trivial solutions (X ) only when A λi (= det(a λi)) = i.e. λ λ = ( λ)[( λ) ] + [ ( λ)] = ( λ)[( λ) ] = λ Evidently, λ = is a root. Factorising out ( λ) leaves ( λ) = = λ = ± = λ =, 3 So the three eigenvalues are λ =, λ = and λ 3 = 3. Calculate the eigenvectors: For eigenvalue λ and eigenvector X = A X = λx X X X 3 we have which gives X +X 3 = λx () X = λx () X +X 3 = λx 3 (3) When λ = λ =, () = X =, () or (3) = X = X 3. The eigenvector has an unspecified magnitude; if X satisfies A X = λx, the ax also satisfies this equation (a = constant). Thus we may take X =

8 When λ = λ =, () = X is arbitrary, () = X 3 =, (3) = X =. So X =. (iv) When λ = λ 3 = 3, () = X =, () or (3) = X = X 3. So X 3 =. X X = (,, ) (,, ) = X X 3 = (,, ) (,, ) = X X 3 = (,, ) (,, ) = i.e. the three eigenvectors are mutually orthogonal (perpendicular) to each other. This is to be expected since A is real and symmetric. = α +α 3 B = α X + α X = α 3 X 3 = = α = α =, α =, α 3 = 3. 4 = α +α 3 (v) C = A B = A (α X + α X + α 3 X 3 ) = α A X + α A X + α 3 A X 3 = α λ X + α λ X + α 3 λ 3 X 3 = ( )() + ()() = 8 4 = C from part (i). + (3)(3) The eigenvectors of a real symmetric matrix are complete, orthogonal and the corresponding eigenvalues are real. (Complete means that any vector may be written as a linear sum of the eigenvectors.)