Question

A horizontal spring has one end fixed and one end attached to a 3 kg mass, which slides without friction. The spring has a stiffness constant of 48 N/m. At time t = 0 the mass is at rest at the origin when a driving force of F = 120 cos 6t (F is in newtons) is applied.

(a) Find the natural oscillation frequency, ω◦, and show that the homogeneous solution (i.e., the solution to the motion when F = 0 ) is xh(t) = A cos 4t + B sin 4t.

(b) Use the method of undetermined coefficients to find the particular solution. Use an initial guess of xp(t) = C cos 6t + D sin 6t, then find values for C and D. [Ans: xp(t) = −2 cos 6t]

(c) Apply the initial conditions to show that the general solution is x(t) = 2 [cos 4t − cos 6t].

(d) Use a trig identity to re-write the general solution in part (c) as x(t) = 4 sin t sin 5t

Answer 1

(a)

_o = (k / m)

= (48 / 3)

= 4 rad / s

suppose x₁(t) and x₁(t) are both solutions of the simple harmonic oscillator equation

(d² / dt²) x₁(t) = - (k / m) x₁(t)

(d² / dt²) x₂(t) = - (k / m) x₂(t)

then the sum

x(t) = x₁(t) + x₂(t) of the two solutions is also a solution to see this consider

(d²x(t) / dt²) = d² / dt²) [x₁(t) + x₂(t)] = (d²x₁(t) / dt²) + (d²x₂(t) / dt²)

using the fact that x₁(t) and x₂(t) both solve the simple harmonic oscillator equation we see that

(d² / dt²) x(t) = - (k / m) x₁(t) + - (k / m) x₂(t)

= - (k / m)[x₁(t) + x₂(t)]

= - (k / m)x(t)

thus the linear combination x(t) =  x₁(t) + x₂(t) is also a solution of the SHO equation

so the sum of sine and cosine solutions is the general solution

x(t) = A cos(_ot) + B sin(_ot) .............(C)

= A cos 4t + B sin 4t

v_x(t) = dx / dt

= - (_o) A sin(_ot) + (_o) B cos(_ot)..............(D)

(b)

to get constants A and B subatitute t = 0 in the above equations C and D so that

A = x_o and B = v_xo / _o

(c)

x(t) = 2 [cos 4t − cos 6t]