TEST BANK FOR The Physics of Vibrations and Waves 6th Edition By H.J. Pain (Textbook author), Youfang Hu

Exam (elaborations) TEST BANK FOR The Physics of Vibrations and Waves 6th Edition By H.J. Pain (Textbook author), Youfang Hu SOLUTIONS TO CHAPTER 1 1.1 In Figure 1.1(a), the restoring force is given by: F = −mg sinθ By substitution of relation sinθ = x l into the above equation, we have: F = −mg x l so the stiffness is given by: s = − F x = mg l so we have the frequency given by: ω2 = s m = g l Since θ is a very small angle, i.e. θ = sinθ = x l , or x = lθ , we have the restoring force given by: F = −mgθ Now, the equation of motion using angular displacement θ can by derived from Newton’s second law: F = m&x& i.e. −mgθ = mlθ&& i.e. θ + θ = 0 l && g which shows the frequency is given by: ω 2 = g l In Figure 1.1(b), restoring couple is given by −Cθ , which has relation to moment of inertia I given by: −Cθ = Iθ&& i.e. θ + θ = 0 I && C which shows the frequency is given by: ω2 = C I © 2008 John Wiley & Sons, Ltd In Figure 1.1(d), the restoring force is given by: F = −2T x l so Newton’s second law gives: F = m&x& = −2Tx l i.e. &x&+ 2Tx lm = 0 which shows the frequency is given by: lm ω2 = 2T In Figure 1.1(e), the displacement for liquid with a height of x has a displacement of x 2 and a mass of ρAx , so the stiffness is given by: Ag x Axg x s G ρ ρ 2 2 2 = = = Newton’s second law gives: −G = m&x& i.e. − 2ρAxg = ρAl&x& i.e. + 2 x = 0 l &x& g which show the frequency is given by: ω 2 = 2g l In Figure 1.1(f), by taking logarithms of equation pVγ = constant , we have: ln p +γ lnV = constant so we have: + = 0 V dV p dp γ i.e. V dp = −γp dV The change of volume is given by dV = Ax , so we have: V dp = −γp Ax The gas in the flask neck has a mass of ρAl , so Newton’s second law gives: Adp = m&x& © 2008 John Wiley & Sons, Ltd i.e. Alx V −γp A x = ρ && 2 i.e. + x = 0 l V x pA ρ γ && which show the frequency is given by: l V pA ρ γ ω 2 = In Figure 1.1 (g), the volume of liquid displaced is Ax , so the restoring force is − ρgAx . Then, Newton’s second law gives: F = −ρgAx = m&x& i.e. + x = 0 m x g A ρ && which shows the frequency is given by: ω 2 = gρA m 1.2 Write solution x = a cos(ωt +φ ) in form: x = a cosφ cosωt − asinφ sinωt and compare with equation (1.2) we find: A = a cosφ and B = −asinφ . We can also find, with the same analysis, that the values of A and B for solution x = asin(ωt −φ ) are given by: A = −asinφ and B = acosφ , and for solution x = acos(ωt −φ ) are given by: A = a cosφ and B = asinφ . Try solution x = a cos(ωt +φ ) in expression &x&+ω 2x , we have: &x&+ω 2x = −aω 2 cos(ωt +φ ) +ω2a cos(ωt +φ ) = 0 Try solution x = asin(ωt −φ ) in expression &x&+ω 2x , we have: &x&+ω 2x = −aω2 sin(ωt −φ ) +ω 2asin(ωt −φ ) = 0 Try solution x = a cos(ωt −φ ) in expression &x&+ω 2x , we have: &x&+ω 2x = −aω 2 cos(ωt −φ ) +ω2a cos(ωt −φ ) = 0 © 2008 John Wiley & Sons, Ltd 1.3 (a) If the solution x = asin(ωt +φ ) satisfies x = a at t = 0 , then, x = asinφ = a i.e. φ =π 2 . When the pendulum swings to the position x = + a 2 for the first time after release, the value of ωt is the minimum solution of equation asin(ωt +π 2) = + a 2 , i.e. ωt =π 4 . Similarly, we can find: for x = a 2 , ωt =π 3 and for x = 0 , ωt =π 2 . If the solution x = acos(ωt +φ ) satisfies x = a at t = 0 , then, x = a cosφ = a i.e. φ = 0 . When the pendulum swings to the position x = + a 2 for the first time after release, the value of ωt is the minimum solution of equation acosωt = + a 2 , i.e. ωt =π 4 . Similarly, we can find: for x = a 2 , ωt =π 3 and for x = 0, ωt =π 2 . If the solution x = asin(ωt −φ ) satisfies x = a at t = 0 , then, x = asin(−φ ) = a i.e. φ = −π 2 . When the pendulum swings to the position x = + a 2 for the first time after release, the value of ωt is the minimum solution of equation asin(ωt +π 2) = + a 2 , i.e. ωt =π 4 . Similarly, we can find: for x = a 2 , ωt =π 3 and for x = 0 , ωt =π 2 . If the solution x = a cos(ωt −φ ) satisfies x = a at t = 0 , then, x = a cos(−φ ) = a i.e. φ = 0 . When the pendulum swings to the position x = + a 2 for the first time after release, the value of ωt is the minimum solution of equation acosωt = + a 2 , i.e. ωt =π 4 . Similarly, we can find: for x = a 2 , ωt =π 3 and for x = 0 , ωt =π 2 . (b) If the solution x = asin(ωt +φ ) satisfies x = −a at t = 0 , then, x = asinφ = −a i.e. φ = −π 2 . When the pendulum swings to the position © 2008 John Wiley & Sons, Ltd x = + a 2 for the first time after release, the value of ωt is the minimum solution of equation asin(ωt −π 2) = + a 2 , i.e. ωt = 3π 4 . Similarly, we can find: for x = a 2 , ωt = 2π 3 and for x = 0 , ωt =π 2 . If the solution x = acos(ωt +φ ) satisfies x = −a at t = 0 , then, x = a cosφ = −a i.e. φ =π . When the pendulum swings to the position x = + a 2 for the first time after release, the value of ωt is the minimum solution of equation acos(ωt +π ) = + a 2 , i.e. ωt = 3π 4 . Similarly, we can find: for x = a 2 , ωt = 2π 3 and for x = 0 , ωt =π 2 . If the solution x = asin(ωt −φ ) satisfies x = −a at t = 0 , then, x = asin(−φ ) = −a i.e. φ =π 2 . When the pendulum swings to the position x = + a 2 for the first time after release, the value of ωt is the minimum solution of equation asin(ωt −π 2) = + a 2 , i.e. ωt = 3π 4 . Similarly, we can find: for x = a 2 , ωt = 2π 3 and for x = 0 , ωt =π 2 . If the solution x = a cos(ωt −φ ) satisfies x = −a at t = 0 , then, x = acos(−φ ) = −a i.e. φ =π . When the pendulum swings to the position x = + a 2 for the first time after release, the value of ωt is the minimum solution of equation acos(ωt −π ) = + a 2 , i.e. ωt = 3π 4 . Similarly, we can find: for x = a 2 , ωt = 2π 3 and for x = 0 , ωt =π 2 . 1.4 The frequency of such a simple harmonic motion is given by: 4.5 10 [ ] 4 8.85 10 (0.05 10 ) 9.1 10 (1.6 10 ) 4 16 1 12 9 3 31 19 2 3 0 2 0 − − − − − ≈ × ⋅ × × × × × × × × = = = rad s r m e m s e e πε π ω Its radiation generates an electromagnetic wave with a wavelength λ given by: © 2008 John Wiley & Sons, Ltd 4.2 10 [ ] 42[ ] 4.5 10 2 2 3 10 8 16 8 0 c ≈ × m = nm × × × × = = − π ω π λ Therefore such a radiation is found in X-ray region of electromagnetic spectrum. 1.5 (a) If the mass m is displaced a distance of x from its equilibrium position, either the upper or the lower string has an extension of x 2 . So, the restoring force of the mass is given by: F = −sx 2 and the stiffness of the system is given by: s′ = −F x = s 2 . Hence the frequency is given by s m s m a ω2 = ′ = 2 . (b) The frequency of the system is given by: s m b ω2 = (c) If the mass m is displaced a distance of x from its equilibrium position, the restoring force of the mass is given by: F = −sx − sx = −2sx and the stiffness of the system is given by: s′ = −F x = 2s . Hence the frequency is given by s m s m c ω2 = ′ =2 . Therefore, we have the relation: 2 : 2 : 2 = s 2m: s m: 2s m = 1: 2 : 4 a b c ω ω ω 1.6 At time t = 0 , 0 x = x gives: 0 asinφ = x (1.6.1) 0 x& = v gives: 0 aω cosφ = v (1.6.2) From (1.6.1) and (1.6.2), we have 0 0 tanφ =ωx v and 2 2 1 2 0 2 0 a = (x + v ω ) 1.7 The equation of this simple harmonic motion can be written as: x = asin(ωt +φ ) . The time spent in moving from x to x + dx is given by: t dt = dx v , where t v is the velocity of the particle at time t and is given by: v = x = aω cos(ωt +φ ) t & . © 2008 John Wiley & Sons, Ltd Noting that the particle will appear twice between x and x + dx within one period of oscillation. We have the probability η of finding it between x to x + dx given by: T η = 2dt where the period is given by: ω 2π T = , so we have: 2 cos( ) cos( ) 1 sin2 ( ) 2 2 2 2 a x dx a t dx a t dx a t dx T dt − = − + = + = + = = π ω ω φ π ω φ π ω φ π ω η 1.8 Since the displacements of the equally spaced oscillators in y direction is a sine curve, the phase difference δφ between two oscillators a distance x apart given is proportional to the phase difference 2π between two oscillators a distance λ apart by: δφ 2π = x λ , i.e. δφ = 2πx λ . 1.9 The mass loses contact with the platform when the system is moving downwards and the acceleration of the platform equals the acceleration of gravity. The acceleration of a simple harmonic vibration can be written as: a = Aω 2 sin(ωt +φ ) , where A is the amplitude, ω is the angular frequency and φ is the initial phase. So we have: Aω 2 sin(ωt +φ ) = g i.e. ω 2 sin(ω +φ ) = t A g Therefore, the minimum amplitude, which makes the mass lose contact with the platform, is given by: 0.01[ ] 4 5 9.8 min 2 4 2 2 2 2 m f A g g ≈ × × = = = ω π π 1.10 The mass of the element dy is given by: m′ = mdy l . The velocity of an element dy of its length is proportional to its distance y from the fixed end of the spring, and is given by: v′ = yv l . where v is the velocity of the element at the other end of the spring, i.e. the velocity of the suspended mass M . Hence we have the kinetic energy © 2008 John Wiley & Sons, Ltd of this element given by: 2 2 2 1 2 1 ⎟⎠ ⎞ ⎜⎝ ⎛ ⎟⎠ ⎞ ⎜⎝ = ′ ′ = ⎛ v l dy y l KE m v m dy The total kinetic energy of the spring is given by: ∫ ∫ ∫ = = ⎟⎠ ⎞ ⎜⎝ ⎛ ⎟⎠ ⎞ ⎜⎝ = = ⎛ l l l spring dy y dy mv l v mv l dy y l KE KE dy m 0 0 2 0 2 3 2 2 6 1 2 2 1 The total kinetic energy of the system is the sum of kinetic energies of the spring and the suspended mass, and is given by: 2 2 ( 3) 2 2 1 2 1 6 KE 1 mv Mv M m v tot = + = + which shows the system is equivalent to a spring with zero mass with a mass of M + m 3 suspended at the end. Therefore, the frequency of the oscillation system is given by: 3 2 M m s + ω = 1.11 In Figure 1.1(a), the restoring force of the simple pendulum is −mg sinθ , then, the stiffness is given by: s = mg sinθ x = mg l . So the energy is given by: 2 2 2 2 2 1 2 1 2 1 2 1 x l E = mv + sx = mx& + mg The equation of motion is by setting dE dt = 0 , i.e.: 0 2 1 2 1 2 2 = ⎟⎠ ⎞ ⎜⎝ ⎛ + x l mx mg dt d & i.e. + x = 0 l &x& g In Figure 1.1(b), the displacement is the rotation angle θ , the mass is replaced by the moment of inertia I of the disc and the stiffness by the restoring couple C of the wire. So the energy is given by: 2 2 2 1 2 E = 1 Iθ& + Cθ The equation of motion is by setting dE dt = 0 , i.e.: 0 2 1 2 1 2 2 = ⎟⎠ ⎞ ⎜⎝ ⎛ Iθ + Cθ dt d & © 2008 John Wiley & Sons, Ltd i.e. θ + θ = 0 I && C In Figure 1.1(c), the energy is directly given by: 2 2 2 1 2 E = 1 mv + sx The equation of motion is by setting dE dt = 0 , i.e.: 0 2 1 2 1 2 2 = ⎟⎠ ⎞ ⎜⎝ ⎛ mx + sx dt d & i.e. + x = 0 m &x& s In Figure 1.1(c), the restoring force is given by: − 2Tx l , then the stiffness is given by: s = 2T l . So the energy is given by: 2 2 2 2 2 2 2 2 1 2 1 2 1 2 1 2 1 x l x mx T l E = mv + sx = mx& + T = & + The equation of motion is by setting dE dt = 0 , i.e.: 0 2 1 2 2 = ⎟⎠ ⎞ ⎜⎝ ⎛ + x l mx T dt d & i.e. + 2 x = 0 lm &x& T In Figure 1.1(e), the liquid of a volume of ρAl is displaced from equilibrium position by a distance of l 2 , so the stiffness of the system is given by s = 2ρgAl l = 2ρgA. So the energy is given by: 2 2 2 2 2 2 2 2 1 2 1 2 1 2 1 2 E = 1 mv + sx = ρAlx& + ρgAx = ρAlx& + ρgAx The equation of motion is by setting dE dt = 0 , i.e.: 0 2 1 2 2 = ⎟⎠ ⎞ ⎜⎝ ⎛ Alx + gAx dt d ρ & ρ i.e. + 2 x = 0 l &x& g © 2008 John Wiley & Sons, Ltd In Figure 1.1(f), the gas of a mass of ρAl is displaced from equilibrium position by a distance of x and causes a pressure change of dp = −γpAx V , then, the stiffness of the system is given by s = − Adp x =γpA2 V . So the energy is given by: V E mv sx Alx pA x 2 2 2 2 2 2 1 2 1 2 1 2 1 γ = + = ρ & + The equation of motion is by setting dE dt = 0 , i.e.: 0 2 1 2 1 2 2 2 = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ + V Alx pA x dt d γ ρ & i.e. + x = 0 l V x pA ρ γ && In Figure 1.1(g), the restoring force of the hydrometer is − ρgAx , then the stiffness of the system is given by s = ρgAx x = ρgA. So the energy is given by: 2 2 2 2 2 1 2 1 2 1 2 E = 1 mv + sx = mx& + ρgAx The equation of motion is by setting dE dt = 0 , i.e.: 0 2 1 2 1 2 2 = ⎟⎠ ⎞ ⎜⎝ ⎛ mx + gAx dt d & ρ i.e. + x = 0 m x A g ρ && 1.12 The displacement of the simple harmonic oscillator is given by: x = asinωt (1.12.1) so the velocity is given by: x& = aω cosωt (1.12.2) From (1.12.1) and (1.12.2), we can eliminate t and get: sin2 cos2 1 2 2 2 2 2 + = t + t = a x a x ω ω ω & (1.12.3) which is an ellipse equation of points (x, x&) . The energy of the simple harmonic oscillator is given by: © 2008 John Wiley & Sons, Ltd 2 2 2 1 2 E = 1 mx& + sx (1.12.4) Write (1.12.3) in form x&2 =ω2 (a2 − x2 ) and substitute into (1.12.4), then we have: 2 2 2 2 2 2 2 ( ) 1 2 1 2 1 2 E = 1 mx& + sx = mω a − x + sx Noting that the frequency ω is given by: ω 2 = s m , we have: 2 2 2 2 2 1 2 ( ) 1 2 E = 1 s a − x + sx = sa which is a constant value. 1.13 The equations of the two simple harmonic oscillations can be written as: sin( ) 1 y = a ωt +φ and sin( ) 2 y = a ωt +φ +δ The resulting superposition amplitude is given by: [sin( ) sin( )] 2 sin( 2)cos( 2) 1 2 R = y + y = a ωt +φ + ωt +φ +δ = a ωt +φ +δ δ and the intensity is given by: I = R2 = 4a2 cos2 (δ 2)sin2 (ωt +φ +δ 2) i.e. I ∝ 4a2 cos2 (δ 2) Noting that sin2 (ωt +φ +δ 2) varies between 0 and 1, we have: 0 ≤ I ≤ 4a2 cos2 (δ 2) 1.14 2 cos( ) (sin cos ) (sin cos ) 2 (sin sin cos cos ) sin sin 2 sin sin cos cos 2 cos cos sin sin cos cos 1 2 1 2 2 2 2 2 1 2 1 2 1 2 1 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 1 2 2 2 2 1 2 1 2 2 2 2 1 2 1 2 1 2 2 2 2 2 2 2 1 2 2 2 1 1 2 2 1 2 2 1 φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ = + − − = + + + − + = + − + + − ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ − + ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ − a a xy a y a x a a xy a y a x a a xy a x a y a a xy a y a x a x a y a y a x On the other hand, by substitution of : 1 1 1 sinωt cosφ cosωt sinφ a x = + © 2008 John Wiley & Sons, Ltd 2 2 2 sinωt cosφ cosωt sinφ a y = + into expression 2 2 1 1 2 2 1 2 2 1 cos cos sin sin ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ − + ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ φ − φ φ φ a x a y a y a x , we have: sin ( ) (sin cos )sin ( ) sin (sin cos sin cos ) cos (cos sin cos sin ) sin sin cos cos 2 1 2 2 1 2 2 2 2 1 2 2 1 2 2 2 1 1 2 2 2 2 1 1 2 2 1 2 2 1 φ φ ω ω φ φ ω φ φ φ φ ω φ φ φ φ φ φ φ φ = − = + − = − + − ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ − + ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ − t t t t a x a y a y a x From the above derivation, we have: 2 cos( ) sin ( ) 2 1 2 1 2 1 2 2 2 2 2 1 2 + − φ −φ = φ −φ a a xy a y a x 1.15 By elimination of t from equation x = asinωt and y = bcosωt , we have: 1 2 2 2 2 + = b y a x which shows the particle follows an elliptical path. The energy at any position of x , y on the ellipse is given by: ( ) 2 1 2 1 2 1 cos 2 sin 1 2 sin 1 2 cos 1 2 1 2 1 2 1 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 m a b ma mb ma t ma t mb t mb t E mx sx my sy = + = + = + + + = + + + ω ω ω ω ω ω ω ω ω ω ω & & The value of the energy shows it is a constant and equal to the sum of the separate energies of the simple harmonic vibrations in x direction given by 2 2 2 1 mω a and in y direction given by 2 2 2 1 mω b . At any position of x , y on the ellipse, the expression of m(xy& − yx&) can be written as: © 2008 John Wiley & Sons, Ltd m(xy& − yx&) = m(−abω sin2ωt − abω cos2ωt) = −abmω(sin2ωt + cos2ωt) = −abmω which is a constant. The quantity abmω is the angular momentum of the particle. 1.16 All possible paths described by equation 1.3 fall within a rectangle of 1 2a wide and 2 2a high, where 1 max a = x and 2 max a = y , see Figure 1.8. When x = 0 in equation (1.3) the positive value of sin( ) 2 2 1 y = a φ −φ . The value of max 2 y = a . So sin( ) 0 max 2 1 = φ −φ = y y x which defines 2 1 φ −φ . 1.17 In the range 0 ≤φ ≤π , the values of i cosφ are −1 ≤ cos ≤ +1 i φ . For n random values of i φ , statistically there will be n 2 values −1 ≤ cos ≤ 0 i φ and n 2 values 0 ≤ cos ≤ 1 i φ . The positive and negative values will tend to cancel each other and the sum of the n values: cos 0 1 Σ → ≠ = n j i i i φ , similarly cos 0 1 → Σ= n j j φ . i.e. cos cos 0 1 1 Σ Σ → = ≠ = n j j n j i i i φ φ 1.18 The exponential form of the expression: a sinωt + a sin(ωt +δ ) + a sin(ωt + 2δ ) +L+ a sin[ωt + (n −1)δ ] is given by: aeiωt + aei(ωt+δ ) + aei(ωt+2δ ) +L+ aei[ωt+(n−1)δ ] From the analysis in page 28, the above expression can be rearranged as: sin 2 2 sin 2 1 δ ae ω δ nδ i t n ⎥⎦ ⎤ ⎢⎣ ⎡ ⎟⎠ ⎞ ⎜⎝ ⎛ − + with the imaginary part: sin 2 sin 2 2 sin 1 δ δ δ ω n n t a ⎥⎦ ⎤ ⎢⎣ ⎡ ⎟⎠ ⎞ ⎜⎝ ⎛ − + which is the value of the original expression in sine term. © 2008 John Wiley & Sons, Ltd 1.19 From the analysis in page 28, the expression of z can be rearranged as: sin 2 (1 2 ( 1) ) sin 2 δ z aeiωt eiδ ei δ ei n δ aeiωt nδ = + + + − = L The conjugate of z is given by: sin 2 * sin 2 δ z ae−iωt nδ = so we have: sin 2 sin 2 sin 2 sin 2 sin 2 sin 2 2 2 * 2 δ δ δ δ δ zz aeiωt nδ ae iωt n a n = ⋅ − = © 2008 John Wiley & Sons, Ltd SOLUTIONS TO CHAPTER 2 2.1 The system is released from rest, so we know its initial velocity is zero, i.e. 0 0 = t= dt dx (2.1.1) Now, rearrange the expression for the displacement in the form: p q t e p q t x F G e( ) F G ( ) 2 2 − + − − − + + = (2.1.2) Then, substitute (2.1.2) into (2.1.1), we have ( ) ( ) 0 2 2 0 ( ) ( ) 0 = ⎥⎦ ⎤ ⎢⎣ ⎡ − + − − + = − + = − + − − = t p q t p q t t p q F G e p q F G e dt dx i.e. qG = pF (2.1.3) By substitution of the expressions of q and p into equation (2.1.3), we have the ratio given by: (r2 4ms)1 2 r F G − = 2.2 The first and second derivatives of x are given by: (A Bt) e rt m m x B r 2 2 − ⎥⎦ ⎤ ⎢⎣ & = ⎡ − + (A Bt) e rt m m r m x rB 2 2 2 4 − ⎥⎦ ⎤ ⎢⎣ ⎡ && = − + + We can verify the solution by substitution of x , x& and &x& into equation: m&x&+ rx& + sx = 0 then we have equation: ( ) 0 4 2 = + ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ − A Bt m s r which is true for all t, provided the first bracketed term of the above equation is zero, i.e. © 2008 John Wiley & Sons, Ltd 0 4 2 − = m s r i.e. r2 4m2 = s m 2.3 The initial displacement of the system is given by: ( ω ω ) cosφ 1 2 x = e−rt 2m C ei ′t +C e−i ′t = A at t = 0 So: cosφ 1 2 C +C = A (2.3.1) Now let the initial velocity of the system to be: ω ( ω ) ω ( ω ) ω sinφ 2 2 2 2 2 1 i C e A m i C e r m x r r m i t r m i t = − ′ ⎟⎠ ⎞ ⎜⎝ + ⎛− − ′ ⎟⎠ ⎞ ⎜⎝ & = ⎛− + ′ − + ′ − − ′ at t = 0 i.e. cosφ ω ( ) ω sinφ 2 1 2 A i C C A m − r + ′ − = − ′ If r m is very small orφ ≈π 2 , the first term of the above equation approximately equals zero, so we have: sinφ 1 2 C −C = iA (2.3.2) From (2.3.1) and (2.3.2), 1 C and 2 C are given by: ( ) ( ) φ φ φ φ φ φ i i C A i A e C A i A e = − − = = + = 2 2 cos sin 2 2 cos sin 2 1 2.4 Use the relation between current and charge, I = q& , and the voltage equation: