#1**+2 **

Look at the values near t = 0.

At t = 0, cos(0) is +1, and since the graph shows y_{2}(0) is positive, then a_{2} must be positive.

For small values of b_{1}t, sin(b_{1}t) ≈ b_{1}t. For b_{1} positive, this is positive, and since the graph shows y_{1}(t) positive for small values of t, then a_{1} must be positive.

a_{1} and a_{2} represent the amplitudes of y_{1} and y_{2} respectively. Since it is clear from the graph that the amplitude of y_{1} is greater than that of y_{2}, we must have a_{1}>a_{2}.

Summarising: 0 < a_{2 }< a_{1}

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_{.}

Alan
Oct 8, 2017