Slope fields and Euler’s method, question 5

Topic: Slope fields and Euler’s method
Difficulty: Standard
Suggested time: 5 min
Marks: 3
Theory: UBC Math 100 §11.2Key results

A mass on a frictionless spring is $x(t)$ centimetres from equilibrium after $t$ seconds, where

\frac{d^{2}x}{dt^{2}} = -25x, \qquad x(0) = 10, \qquad x'(0) = 50.

Find the positive $\omega$ for which $x(t) = A\cos(\omega t) + B\sin(\omega t)$ is a solution, use the initial conditions to find $A$ and $B$ , and give $x$ at $t = \tfrac{\pi}{20}$ .

Your answer

An exact expression, or a decimal to two decimal places.

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Theory UBC Math 100 §11.2Key results

Worked solution

Strategy

Substitute the proposed form into the equation to find which ω works, then let the two initial conditions fix the two constants, one each.

Working

$x' = -A\omega\sin(\omega t) + B\omega\cos(\omega t), \qquad x'' = -A\omega^{2}\cos(\omega t) - B\omega^{2}\sin(\omega t) = -\omega^{2}x$
Each differentiation brings out a factor ω by the chain rule.
$-\omega^{2}x = -25x \text{ for every } t \iff \omega^{2} = 25 \iff \omega = 5$
Take the positive root, as asked.
$x(0) = A = 10, \qquad x'(0) = 5B = 50 \implies B = 10$
At t = 0, sin 0 = 0 and cos 0 = 1, so each condition fixes one constant.
$x\left(\tfrac{\pi}{20}\right) = 10\cos\tfrac{\pi}{4} + 10\sin\tfrac{\pi}{4} = 10\sqrt{2} \approx 14.14$
Here 5t = π/4, where cos and sin both equal √2/2.