A point moves harmonically along the x-axis with velocity

A point moves harmonically along the $x$-axis with velocity $v=\dfrac{2\pi}{T}\cos\left(\dfrac{2\pi t}{T}+\varphi_0\right)$, where $T$ is the period and $\varphi_0$ the initial phase. Given that the point is at $x=x_1$ at time $t=t_1$, find its position at time $t_2$.

Step 1: Integrate velocity

With $u=\dfrac{2\pi t}{T}+\varphi_0$, $du=\dfrac{2\pi}{T}dt$, so

\[x(t)=\int\frac{2\pi}{T}\cos\!\left(\frac{2\pi t}{T}+\varphi_0\right)dt=\sin\!\left(\frac{2\pi t}{T}+\varphi_0\right)+C.\]

Step 2: Fix the constant using the condition at $t_1$

\[x_1=\sin\!\left(\frac{2\pi t_1}{T}+\varphi_0\right)+C \quad\Longrightarrow\quad C=x_1-\sin\!\left(\frac{2\pi t_1}{T}+\varphi_0\right).\]

Step 3: Position at time $t_2$

\[x_2=\sin\!\left(\frac{2\pi t_2}{T}+\varphi_0\right)+x_1-\sin\!\left(\frac{2\pi t_1}{T}+\varphi_0\right).\]

\[\boxed{x_2=x_1+\sin\!\left(\frac{2\pi t_2}{T}+\varphi_0\right)-\sin\!\left(\frac{2\pi t_1}{T}+\varphi_0\right)}\]

Step 4: Compact form via a sum-to-product identity

Using $\sin A-\sin B=2\cos\!\left(\dfrac{A+B}{2}\right)\sin\!\left(\dfrac{A-B}{2}\right)$ with $A=\dfrac{2\pi t_2}{T}+\varphi_0$, $B=\dfrac{2\pi t_1}{T}+\varphi_0$:

\[\boxed{x_2=x_1+2\cos\!\left(\frac{\pi(t_2+t_1)}{T}+\varphi_0\right)\sin\!\left(\frac{\pi(t_2-t_1)}{T}\right)}\]

Step 5: Consistency check

Setting $t_2=t_1$ in either form immediately gives $x_2=x_1$, as it must.