Question

cos θ cos φ sin φ sin θ, (Beats) Using the trigonometric identities cos(θ verify that φ) (β a) 2 (19) cos ot - cos Bt 2 sin A spring-mass system has an attached mass of 4 g, a spring constant of 16 g/s* and a negligible friction. It is subject to a force of 4 cos(2.2t) down- ward, and is initially 0 at rest. Determine the subsequent motion. Using (19) from Exercise 11, rewrite the solution as the product of two sine functions, and graph the result. The resulting function has a periodic variation in amplitude, or a beat.

Answer 1

sin φ sin θ, (Beats) Using the trigonometric identities cos (θ verify that φ) cos θ cos φ cos cut-cos ßt = 2 sin A spring-mass system has an attached mass of 4 g, a spring constant of 16 g/s2 and a negligible friction. It is subject to a force of 4 cos(2.2t) down- ward, and is initially 0 at rest. Determine the subsequent motion. Using (19) from Exercise 11, rewrite the solution as the product of two sine functions, and graph the result. The resulting function has a periodic variation in amplitude, or a beat. Ans.:

-co (dt) os(at)-sin(a)sin(#)-Lcos(a) cos(a) + sin(θ)sin(#) -cos(0t)cos(#) _ sin (6t)sin(at)-cos (0t) co (φ) _ sin (dt) sin (φ ) -2sin (Ot) sin (pt) 2sin(Ot)sin(-t) sin (Ot sin cos(0t)cos COS 2 2 cos ot-cos β t = 2 sin t sin 2 2

The differential equation for a spring-mass system is given by, at 44x +16x=4c0 s(2.2t).x(0)= 0, x'(0)=0 +4x- cos (2.2) The solution to the homogeneous differential equation--+4x=0is given by, dt G cos 2t+C2 sin2t The particular solution is given by, x, (t)-Acos(2.2r)+Bsin (2.21) x (r) =-2.2.4 sin (2.2t) +2.28 cos (2.2t) x (t)--4.84Acos (2.20-4.84B sin (2.2t) -4.84Acos (2.2)-4.84B sin (2.2t)- ) +4Acos(2.2t)+4B sin (2.2t) -0.84A cos (2.20-0.84B sin (2.2t) = cos (2.2t) = cos ( 2.2t 0.84 (r) =-034cos(2.2t) 0.84

The general solution to the differential equation^+4x- cos (2.2t) is given by, at x(t)=x, (t)+Xy(t)-G cos 2t + C, sin 2t_-cos(2.2t) 0.84 Use the initial conditions x (o)-0,x (o)-0 x(0)-C1 cos0+ C, sin0-0.84cos(0)-04C-014 2.2 x (t)--2C, sin 2t2C, cos2t- sin (2.2t x(0)--2C, sin 0+2C, cos0+sin (0)C0 x(t) = 0.84cos2-0.84 cos(2.2t) 0.84 2.2 0.84 )=014|2sin(22-2t)sin(22+2t) x(t)-034(cos2- cos (2.2t) 0.84 0.84 =045 sin(0.1t)sin(2.1t)

x(t) 2.5 x(t)-sin(0.1t)sin(2.1t) 0.42 1.5 0.5 -4 810 12 416 18 0.5 -1 1.5 -2 2.5