A point moves harmonically, x = a sin kt. Determine

A point moves harmonically, $x = a\sin kt$. Determine the amplitude $a$ and the circular frequency $k$ given that $v = v_{1}$ when $x = x_{1}$, and $v = v_{2}$ when $x = x_{2}$.

Step 1: Eliminate the time

\[ x = a\sin kt, \qquad v = \frac{dx}{dt} = ak\cos kt \;\Longrightarrow\; \sin kt = \frac{x}{a}, \quad \cos kt = \frac{v}{ak}. \]

Squaring and adding removes $t$ entirely, by the Pythagorean identity: \[\frac{x^{2}}{a^{2}}+\frac{v^{2}}{a^{2}k^{2}} = 1 \;\Longrightarrow\; v^{2} = k^{2}\bigl(a^{2}-x^{2}\bigr). \tag{1}\]

Step 2: Two states, two equations

\[ v_{1}^{2} = k^{2}\bigl(a^{2}-x_{1}^{2}\bigr), \qquad v_{2}^{2} = k^{2}\bigl(a^{2}-x_{2}^{2}\bigr). \tag{2} \]

Step 3: Subtract to find $k$

The unknown $a^{2}$ cancels: \[v_{1}^{2}-v_{2}^{2} = k^{2}\bigl(x_{2}^{2}-x_{1}^{2}\bigr) \;\Longrightarrow\; k = \sqrt{\frac{v_{1}^{2}-v_{2}^{2}}{x_{2}^{2}-x_{1}^{2}}}. \tag{3}\]

Step 4: Back-substitute to find $a$

\[ a^{2} = x_{1}^{2}+\frac{v_{1}^{2}}{k^{2}} = x_{1}^{2}+v_{1}^{2}\cdot\frac{x_{2}^{2}-x_{1}^{2}}{v_{1}^{2}-v_{2}^{2}} = \frac{x_{1}^{2}\bigl(v_{1}^{2}-v_{2}^{2}\bigr)+v_{1}^{2}\bigl(x_{2}^{2}-x_{1}^{2}\bigr)}{v_{1}^{2}-v_{2}^{2}}, \]

and the terms $x_{1}^{2}v_{1}^{2}$ cancel in the numerator: $a = \sqrt{\frac{x_{2}^{2}v_{1}^{2}-x_{1}^{2}v_{2}^{2}}{v_{1}^{2}-v_{2}^{2}}}$.

Step 5: Verification

\[\text{Test } a = 2,\ k = 3:\ x_{1} = 0 \Rightarrow v_{1} = 6; \quad x_{2} = 2 \Rightarrow v_{2} = 0\]

\[k = \sqrt{\frac{36-0}{4-0}} = 3 \;\;\checkmark, \qquad a = \sqrt{\frac{4\cdot36-0}{36}} = 2 \;\;\checkmark\]

\[\text{Consistency requires } \bigl(v_{1}^{2}-v_{2}^{2}\bigr) \text{ and } \bigl(x_{2}^{2}-x_{1}^{2}\bigr) \text{ to have the same sign} \;\;\checkmark\]