Answer to Question #164536 in Trigonometry for juls

Question

A weight on the end of a spring oscillating in harmonic motion. The equation model for oscillation is d (t) = 6 sin, 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? 

Answer 1

equation for oscillation is d (t) = 6 sin2t .

a)

period of motion(T) = $\frac{2\pi}{\omega}$ = $\frac{2\pi}{2}$ = $\pi$ sec.

frequency (f) = $\frac{1}{T}$ = $\frac{1}{\pi}$ sec^-1

b)

at equilibrium t= 0, d(0) = 6 sin2(0) = 0

at time t = 2.5, d(2.5) = 6 sin2(2.5) = 6 sin5, here the sin5, angle 5^c is in radian so to convert it into degree we use 360⁰ = (2$\pi$)^c

1^c (radian) = 360/2$\pi$ = 180/$\pi$

thus, d(2.5) = 6*sin( (5*180)/$\pi$ ) = 6*(- 0.959) = -5.75 cm

range of displacement [-6,6] as sin range [-1,1]

thus, the weight is moving away from the equilibrium at t = 2.5 sec.

c)

at t = 3.5 sec , d(3.5) = 6* sin(2*3.5) = 6* sin(7)

d(3.5) = 6* sin((7*180)/$\pi$ ) = 6* (0.665) = 3.991 cm

thus, the weight is moving away from the equilibrium.

d)

at t = 1 sec,

d(1) = 6*sin(2*1)^c = 6* sin((2*180)/$\pi$ )⁰ = 6* 0.909 = 5.455 cm

at t=1.5 sec,

d(1.5) = 6* sin(2*(1.5))^c = 6* sin((3*180)/$\pi$ )⁰ = 6* 0.141 = 0.847 cm

thus, the weight move between t= 1 and t = 1.5 sec is

d(1.5) - d(1) = 0.847 - 5.455 = -4.607 cm

e)

average velocity = $\frac{d(d)}{dt}= (6*cos2t)*{\frac{d(2t)}{dt}} = (6*cos2t)*(2) = 12cos2t$

average velocity at t= 1sec to t = 1.5 sec is

V = 12cos2(1.5) - 12cos2(1) = 12(cos3 - cos2)

V = 12 [cos(3*180/$\pi$ ) - cos(2*180/$\pi$ )] = 12[-0.99 - (-0.42)] = 12*(-0.57)

V = - 6.89 cm/sec.

negative sign shows that the velocity is decreasing as time increase from 1 sec to 1.5 sec.