(a)A mass weighing 4 lb stretches a spring 1.5 in. The mass is displaced 2 in. in the positive
direction from its equilibrium position and released with no initial velocity. Assuming that
there is no damping and that the mass is acted on by an external force of 2 cos 3t lb,
formulate the initial value problem describing the motion of the mass.
(b) Find the solution of the above IVP
(c) If the given external force is replaced by a force $4 sin ωt$ of frequency $ω$, find the value
of $ω$ for which resonance occurs.
What i tried
(a)Since this is a Forced undamped Oscillation, the IVP should be of the form $$mu''+ku=F(t)$$
Hence $m$ is the mass 4 Ib
$L=1.5/12=0.125$
While $$k=mg/L=(4)(32)/0.125=1024 $$
$$F(t)=2 cos 3t $$
Hence the IVP i got is $$4u''+1024u=2cos(3t)$$ $$u(0)=2/12$$ $$u'(0)=0$$
(b)The above iVP will have complex solutions hence
$$W_o=\sqrt{k/m}$$, hence this gives a value of $16$
And the complementary solution will be $$y(t)=C_{1}cos(16t)+C_{2}sin(16t)$$
while the method of undetermined coefficients is used to find the particular solutions. Am i correct and also im unsure of how to continue to part (c). COuld anyone please explain. Also while im famillar with the metric system, im quite confused when the question uses non metric system. such as in this question. Could anyone also explain the conversion and use for the non metric system espically for this problem. Thanks
Best Answer
i am assuming you have reached $$u'' + 256u = \cos \omega t $$ and you are looking for the resonance frequency $\omega.$ trying a particular solution(response) to the unit forcing with frequency $\omega$ of the form $$u = A\cos \omega t $$ we find that the amplitude is $$A = \frac1{256 - \omega^2}$$ which goes unbounded as $\omega \to 16.$ therefore $\omega = 16$ is the resonance frequency.
You can determine the form of the solution it takes at resonance by looking at a particular solution, when $\omega$ is close to $16,$ of the form $$u = \left(\frac{\cos \omega t - \cos 16 t}{256 - \omega^2}\right) = -\frac{\cos \omega t - \cos 16 t}{(\omega + 16 )(\omega - 16 )} \to -\frac{1}{32} \frac{d\cos \omega t}{d \omega}\Big|_{\omega=16} =\frac{1}{32} t \sin 16 t $$