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)$$ Equation (4) can be broken into two parts it's amplitude (if I am correct??) of $\frac{2}{\omega_0^2-\omega_1^2}\frac{\sin(\omega_0-\omega_1)}{2}t$ and oscillation of $\sin\omega_0t$?? And it should also equal equation (2) so $\frac{t}{2\omega_0}$ from equation (2) equals $\frac{2}{\omega_0^2-\omega_1^2}\frac{\sin(\omega_0-\omega_1)}{2}t$ from equation (4) but they don't seem to agree? Appreciate any insight you could offer!
Simple Harmonic Motion with forcing
calculusordinary differential equationsphysics
Related Solutions
If the ratio $r=\dfrac{\omega_1}{\omega_2}$ is not rational, function $f$ can reach values arbitrarily close to $|A_1|+|A_2|$.
A graphical example in the case $f(t)=2 \cos(1.5t)-3 \cos(\sqrt{2}t)$ with values arbitrarily close to $|2|+|-3|=5$.
A sketch of proof: As remarked by Yves Daoust, the case $A_1>0$ and $A_2>0$ is evident. Let us consider, for the sake of definiteness, the case $A_1>0$ and $A_2<0$.
It suffices to find a value of $t$ simultaneously:
very close to $\dfrac{2k\pi}{\omega_1}$, giving a value of the first cosine is very close to $1$, and
very close to $\dfrac{(2\ell+1)\pi}{\omega_2}$, so as the second cosine is very close to $-1$.
(with with $k,\ell \in \mathbb{Z}$). These approximate identities:
$$t \approx \dfrac{2k\pi}{\omega_1} \approx \dfrac{(2\ell+1)\pi}{\omega_2} \ \ \ (1)$$
can be reinterpreted by saying that it suffices to be able to find integers $k$ and $\ell$ such that:
$$\dfrac{\omega_1}{\omega_2}\approx\dfrac{2k}{2\ell+1}$$
otherwise said, to be able to find a "very good" rational approximation of $r=\dfrac{\omega_1}{\omega_2}$ of the form $\dfrac{2k}{2\ell+1}$. This is a consequence of the density of the rationals in the reals (even of the rationals of the form $\dfrac{2k}{2\ell+1}$).
First off, determine how many independent dimensionless quantities there are. You have three independent dimensions and eight quantities (six parameters, time, and the angle), so you have five independent dimensionless quantities.
Naturally one of them is just $\theta$.
Since you have some natural frequencies of oscillation, it makes sense to have another dimensionless quantity be either $\omega t$ (with $\omega$ the natural frequency of the unforced, undamped linear pendulum) or $\Omega t$. You seem to have chosen it to be $\omega t$.
It is also natural to have a dimensionless parameter be a nondimensionalized version of $\omega-\Omega$ because you are specifically interested in beating. You wrote this as $\frac{\omega-\Omega}{\omega}$. This makes sense.
Now we need two more. The setup you're going for here compares the external forcing to the gravitational forcing, and the damping to the gravitational forcing. This is a choice, although it is a pretty reasonable one.
But you need the right units for $k$. The way you wrote it, the units for $k$ are $M L^2 T^{-1}$. This is a slightly weird convention (different from the convention for the damping coefficient in the usual spring-mass system with linear damping). But it doesn't hurt anything as long as you're aware of it.
Anyway, since the only source of a time that you're "allowed" to use in this term is $g$, you divide by $\sqrt{g}$ to cancel $T^{-1}$ and are now looking at $\frac{k}{\sqrt{g}}$, which has $ML^{3/2}$ units. Then you cancel what's left over with the simplest thing that has $ML^{3/2}$ units available which is $m L^{3/2}$.
Best Answer
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 $\frac t{2\omega_0}$.