Find Amplitude of each period of time series
I assume from your solution that your time series closely fits a sine wave.  The method you are using is quite good if you restrict yourself to the set of data which is near the peaks.

The more standard way is to find a best fit sinusoid using linear algebra techniques.  You have stated the period, so you know the frequency.  Let's assume you don't know where the zero crossings are.  The problem then becomes to find the best values of $(a,b)$ so that

$$ Y = a C + b S $$

Where $Y$ is your signal values as a vector, $C$ is a cosine curve, and $S$ is a sine curve over one period.  Now dot this equation with $C$ and then $S$ to get:

$$ Y \cdot C = a C \cdot C + b S \cdot C $$
$$ Y \cdot S = a C \cdot S + b S \cdot S $$

The dot products are scalars.  Since $S$ and $C$ are orthogonal, their dot product is zero.  The dot product of $S$ or $C$ with itself is $N/2$ where $N$ is the sample count.  The equations then become

$$ Y \cdot C = a \cdot N/2 $$
$$ Y \cdot S = b \cdot N/2 $$

Solve for $a$ and $b$.

$$ a = \frac{ Y \cdot C }{ N/2 } $$
$$ b = \frac{ Y \cdot S }{ N/2 } $$

Your amplitude can now be found from $a$ and $b$.

$$ A = \sqrt{a^2+b^2} $$

This method is equivalent to a single bin of a Discrete Fourier Transform (DFT), so if you follow this then you understand how a DFT works.

To calculate the dot product, you take a summation:

$$ Y \cdot C = \sum_{n=0}^{N-1} { Y[n] C[n] } $$
$$ Y \cdot S = \sum_{n=0}^{N-1} { Y[n] S[n] } $$

That's all there is to it.