Question

Show that simple harmonic motion y(t) = C1 cos ωt + C2 sin ωt can be written as:

a. y(t) = A sin(ωt + ϕ0)

b. y(t) = A cos(ωt + ϕ1)

Answer 1

At first, rewrite the function in the following way: $y(t)=\sqrt{C_1^2+C_2^2}(\frac{C_1}{\sqrt{C_1^2+C_2^2}}\cos\.wt+\frac{C_2}{\sqrt{C_1^2+C_2^2}}\sin\.wt)$. Point out that $-1\leq\frac{C_1}{\sqrt{C_1^2+C_2^2}}\leq1$ and $\left(\frac{C_1}{\sqrt{C_1^2+C_2^2}}\right)^2+\left(\frac{C_2}{\sqrt{C_1^2+C_2^2}}\right)^2=1$ Therefore, there is such $\phi_0$ that $\frac{C_1}{\sqrt{C_1^2+C_2^2}}=\cos\phi_0$ and $\frac{C_2}{\sqrt{C_1^2+C_2^2}}=\sin\phi_0$. It can be obtained from the properties of functions $\arccos$, $\arcsin$. We receive: $y(t)=(\cos\phi_0\,\cos\.wt+\sin\phi_0\sin\.wt)=\cos(wt-\phi_0)$. Thus, we get: $y(t)=A\,\cos\.(wt-\phi_0)$, where $A=\sqrt{C_1^2+C_2^2}$ . After the change: $\phi_0\rightarrow-\phi_1$ we receive formula b). After the change $w\rightarrow-w$, $\phi_0\rightarrow\phi_0-\frac{\pi}2$ in formula $y(t)=A\,\cos\.(wt-\phi_0)$ we receive: $y(t)=A\,\cos\.(\frac{\pi}2-wt-\phi_0)=A\sin(wt+\phi_0)$. It is formula a). .