A spring withk = 35.5 N/m has a mass of 5.50 kg attached to it. An external force F = (4.65 N)sin[(6.80 s¯)t] drives the spring mass system so that it oscillates without any resistive forces. What is the amplitude of theSoscillatory motion of the spring-mass system?

This case is for the forced oscillations. In this case, we do not have a resistive force. F(t) Let us consider the case where the mass of the block be m kg and the spring constant be k Nm and an external force acting on the system be F0 sinωt. If we consider damping then if we displace the block by an amount x then the equation of motion for the block is given by md2xdt2=-kx-pdxdt+F(t) Here d2xdt2 is the acceleration of the body. -kx is the restrictive force of the spring and F(t) is the applied force on the system. To find x we have to solve the differential second order inhomogeneous differential equation d2xdt2+kmx+γdxdt=F(t) The solution has two parts. The complementary function is obtained by solving the differential equation d2xdt2+kmx+γdxdt=0 This is done by assuming a trial solution. Then the particular integral is found by the D operator method. Let the solution of this equation has a form x=B cos(ωt-ϕ) We write this in complex form as x=B ei(ωt-ϕ) We also write the force in the form of F(t)=F0meiωt Therefore substituting the value of x in the differential equation we get d2xdt2+ω02x+γdxdt=F0meiωt=F0meiωt-ϕ+ϕ=F0meiωt-ϕeiϕ Substituting the value of x in the differential equation we get Bω02-ω2+iωγ-F0meiϕ=0 ⇒B=F0meiϕω02-ω2+iωγ⇒B e-iϕ=F0meiϕω02-ω2+iωγ⇒Bcosϕ-isinϕ=F0meiϕω02-ω2+iωγ Taking complex conjugate of above equation and adding and subtracting we get B cosϕ=F0mω02-ω2ω02-ω22+ω2γ2B sinϕ=ωγF0mω02-ω22+ω2γ2 Now taking square on both the sides and adding we get B=F0m{ω02-ω22+ω2γ2}1/2 In our case we have γ=0 then B=F0mω02-ω2 Now we have in the question k=35.5 N/m and the mass m=5.50 kg. Therefore the differential equation of the mass is given by d2xdt2+35.55.50x=4.65 sin(6.80t) d2xdt2+6.45x=4.65 sin(6.80t) Now we have ω02=km=6.45 and ω2=6.802=46.24. Therefore we have Amplitude B=4.655.5046.24-6.45=0.02124 m Answer: The amplitude of the oscillation is 0.02124 m

Answered: A spring with k = 35.5 N/m has a mass…

A spring with k = 35.5 N/m has a mass of 5.50 kg attached to it. An external force F = (4.65 N)sin[(6.80 s-)t] drives the spring mass system so that it oscillates without any resistive forces. What is the amplitude of the oscillatory motion of the spring-mass system?