Answer to Question #89862 in Mechanics | Relativity for gerald

Question

the motion of a particle is describe by x=10sin2t + 8cos2t. determine the period, amplitude and phase angle

Answer 1

$$x = 10 \sin 2t + 8 \cos 2t$$ $$x = 2 \sqrt{4^2 + 5^2} \bigg(\frac{5}{\sqrt{4^2 + 5^2}} \sin 2t + \frac{4}{\sqrt{4^2 + 5^2}} \cos 2t\bigg)$$

Let us denote

$$\cos \phi = \frac{5}{\sqrt{4^2+5^2}}$$

then

$$\sin \phi = \sqrt{1-\cos^2 \phi} = \sqrt{1 - \frac{5^2}{4^2 - 5^2}} = \frac{4}{\sqrt{4^2 + 5^2}}$$

and

$$\phi = \arctan \frac{4}{5}$$

Using

$$\cos \phi \sin 2t + \sin \phi \cos 2t = \sin (2t+\phi)$$

we obtain

$$x = 2 \sqrt{41} \sin(2t+\phi)$$

The amplitude is

$$2 \sqrt{41}$$

the period is

$$T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi$$

and the phase angle is

$$\phi = \arctan \frac{4}{5} = 0.67 \approx 39^\circ$$