Question

1*. If a mass is attached to a spring raised r, feet and is given an initial vertical velocity of v, feet per second then the susequent position r of the mass is given by ,cos(ut)sin(ut) for time t and postive constant w .) Ifw-2; 2% = 2 feet; and vo-1 ft/sec write r in the form rsin(wt +-5). Give δ in radians rounded to the nearest throusanth. Find the amplitude and frequency of the spring's oscillation. b.) Determine when the mass passes through the equilibrium position for the first time. u and a, remain constant and v, 3 feet per second. Use the reduction formula to write z in the sim (ut +8). Give 6 in radians rounded to the nearest throusanth. Find the amplitude and frequency of the spring's oscillation. d.) Discuss the effects on amplitude the frequency if v, is less than 1 foot per second. Use complete sente and correct grammar to justify your answer and reasoning

Answer 1

$x=x_0\cos\omega t+\frac{v_0}{\omega}\sin\omega t$

Let

$r=\sqrt{x_0^2+\left ( \frac{v_0}{\omega} \right )^2}$

and

$\sin\delta=\frac{x_0}{r}, \cos\delta=\frac{v_0}{r\omega}$

and thus

$x=r\left (\frac{x_0}{r}\cos\omega t+\frac{v_0}{r\omega}\sin\omega t \right )$

$x=r\left (\sin\delta\cos\omega t+\cos\delta\sin\omega t \right )=r\sin(\omega t+\delta)$

Numerically

(a)

$r=\sqrt{2^2+\left ( \frac{1}{2} \right )^2}=2.0616\textup{ ft}$

and

$\tan\delta=\frac{x_0\omega}{v_0}=4$

$\delta=\tan^{-1}4=1.3258$

Frequency is already given and it is 2 rad./sec

(b)

Equilibrium corresponds to x=0, thus the first solution is

$\omega t+\delta=\pi\Rightarrow t=\frac{\pi-\delta}{\omega}$

$t\approx0.9079\textup{ sec}$

In a similar fashion you can do the other parts. Just change v to 3 into above parts.