Question

A weight on the end of a spring oscillating in harmonic motion. The equation model for oscillation is d (t) = 6 sin ((pi/2)(t)) where d is the distance (in centimeter) from the equilibrium point in t sec.

a. What is the period of the motion? What is the frequency of the motion?

b. What is the placement from the equilibrium at t = 2.5? Is the weight moving toward the equilibrium at this time?

c. What is the displacement from equilibrium at t = 3.5? Is the weight moving toward the equilibrium point or away from the equilibrium at this time?

d. How far does the weight move between t = 1 and t = 1.5 sec?

e. What is the average velocity for this interval? 

f. Do you expect a greater or lesser velocity for t = 1.75 to t = 2? Explain why.

 

Answer 1

$d(t) =6\sin(\frac{\pi}{2}t)$

$a)\omega= \frac{\pi}{2};\omega-\text{ω is the angular frequency}$

$T=\frac{2\pi}{\omega}=4;T\text{ is period of the motion}$

$f=\frac{1}{T}=0.25;f \text{ is the frequency of the motion}$

$b)d(2.5)=6\sin(\frac{\pi}{2}*2.5)\approx-4,24$

$v(t) = d'(t);v(t)\text{velocity object}$

$v(t)=\frac{6\pi}{2}\cos{(\frac{\pi}{2}t)}$

$\text{if }v(t)>0\text{ weight moving from the equilibrium}$

$\text{if }v(t)<0\text{ weight moving toward the equilibrium}$

$v(2.5)=\frac{6\pi}{2}\cos{(\frac{\pi}{2}*2.5)}=-6.6<0$

$\text{then the weight moving toward the equilibrium}$

$c)d(3.5)=6\sin(\frac{\pi}{2}*3.5)\approx-4,24$

$v(3.5)=\frac{6\pi}{2}\cos{(\frac{\pi}{2}*3.5)}=6.6>0$

$\text{then the weight moving from the equilibrium}$

$d)\ d(1)=6\sin(\frac{\pi}{2})\approx=6$

$\ d(1.5)=6\sin(\frac{\pi}{2})\approx4,24$

$e) v_{avg}=\frac{\Delta{s}}{\Delta{t}}=\frac{4.24-6}{1.5-1}=-3.52$

$f)v(t)=\frac{6\pi}{2}\cos{(\frac{\pi}{2}t)}$

$v(1.75)=\frac{6\pi}{2}\cos{(\frac{\pi}{2}*1.75)}\approx-8.7$

$v(2)=\frac{6\pi}{2}\cos{(\frac{\pi}{2}*2)}\approx-9.4$

$|v(2)|>|v(1.75)|$