Simple Harmonic Motion with forcing

Question

Have a few questions on a Simple Harmonic Motion (SHM) it's a swinging pendulum with no damping. Assuming the differential equations that describes it motion as: $$\frac{d^2x}{dt^2} + \omega_0x = \cos\omega_1t (1)$$ where $\omega_0 =$ natural angular frequency $\omega_1 = $ driving\input frequency my question is why use cosine as a driving\input force instead of sine, I understand it has to be sinusoidal but why cosine? My other question is based on a proof I saw on Youtube video and I just wanted to confirm my understanding of the proof so same equation as above $$\frac{d^2x}{dt^2} + \omega_0x = \cos\omega_1t$$ so solution to equation is made up of $X(t) = x_p + x_c$ where $X_p =$ particular solution to differential equation above (1) and $x_c$ is the solution to the homogeneous equation (1) and the driving frequency $\omega_1$ is approximately equal to natural frequency $\omega_0$ so solution becomes $$ x_p = \frac{t\sin\omega_0t}{2\omega_0} (2)$$ would I be right in thinking $\frac{t}{2\omega_0}$ is the amplitude for this solution?$$$$Finally the proof says if $\omega_0$ is approximately equal to $\omega_1$ the solution to the differential equation $\frac{d^2x}{dt^2} + \omega_0x = \cos\omega_1t$ is a linear combination of the particular solution $\frac{\cos\omega_1t}{\omega_0^2-\omega_1^2}$ and complementary solution which is $\frac{\cos\omega_0t}{\omega_0^2-\omega_1^2}$ so $$ x(t) = \frac{\cos\omega_1 -\cos\omega_0t}{\omega_0^2-\omega_1^2} (3) $$ using the trig identity $\cos A - \cos B = 2\sin\frac{A-B}{2}\sin\frac{A+B}{2}$ where A =$\omega_1$ and B = $\omega_0$ so equation (3) becomes $$2\sin\frac{\omega_1-\omega_0}{2}\sin\frac{\omega_1+\omega_0}{2}$$ as $\omega_1$ approx equal $\omega_0$ this equation becomes $$\frac{2}{\omega_0^2-\omega_1^2}\frac{\sin(\omega_0-\omega_1)}{2}t\sin\omega_0t (4)$$ Equation (4) can be broken into two parts it's amplitude (if I am correct??) of $\frac{2}{\omega_0^2-\omega_1^2}\frac{\sin(\omega_0-\omega_1)}{2}t$ and oscillation of $\sin\omega_0t$?? And it should also equal equation (2) so $\frac{t}{2\omega_0}$ from equation (2) equals $\frac{2}{\omega_0^2-\omega_1^2}\frac{\sin(\omega_0-\omega_1)}{2}t$ from equation (4) but they don't seem to agree? Appreciate any insight you could offer!

Answer 1

You are working in the limit $\omega_0\approx\omega_1$. Let's define $\Delta\omega=\omega_0-\omega_1$. Then you can write your amplitude in expression (4) as $$\frac 2{\omega_0^2-\omega_1^2}\frac{\sin(\omega_0-\omega_1)}2 t=\frac{\sin\Delta\omega}{(\omega_0+\omega_1)\Delta\omega}t=\frac{\sin\Delta\omega}{(2\omega_0-\Delta\omega)\Delta\omega}t$$ In the limit $\Delta\omega\to 0$, you have $\frac{\sin\Delta\omega}{\Delta\omega}\to 1$ and $(2\omega_0-\Delta\omega)\to 2\omega_0$. Therefore your amplitude will become $\frac t{2\omega_0}$.