# Simple Harmonic Motion with forcing

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!

• There is no reason why the driving term has to be a cosine. It can be any function. It can be a sine. Whether it is sine or cosine depends on where the timing starts. If the driving force is $0$ at $t=0$ and increasing then you use sine. If the driving force is a maximum at $t=0$ and decreasing then you use cosine. Generally you use $\cos(\omega_t t+\phi)$ and choose $\phi$ appropriately. Commented Apr 11, 2020 at 20:27
• Thanks for the response to my first question Sammy. Commented Apr 12, 2020 at 14:38

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 $$\frac t{2\omega_0}$$.