The 2.1-kg mass is released from rest at a distance x 0 = 55 mm to the right of

Question

The 2.1-kg mass is released from rest at a distance x0 = 55 mm to

the right of the equilibrium position. Determine the displacement x

as a function of time t, where t = 0 is the time of release. After

you have the general expression, answer the questions.

When t = 2.47 s: x=

Answers must be in mm and please claculate and give an answer not just formulas

here is a link to the problem and diagram:

http://i1334.photobucket.com/albums/w655/salonemark/8-32_zpsb29280b5.jpg

When t = 0.31 s: x =
When t = 1.07 s: x =
When t = 1.88 s: x =
The 2.1-kg mass is released from rest at a distance x0 = 55 mm to the right of the equilibrium position. Determine the displacement x as a function of time t, where t = 0 is the time of release. After you have the general expression, answer the questions. When t = 0.31 s: X = When t = 1.07 s: X = When t = 1.88 s: X = When t = 2.47 s: x= Answers must be in mm and please calculate and give an answer not just formulas

Explanation / Answer

a) x = A sin (wt+ phi)

we have amplitude =A= x0 = 55 mm

at t=0,

we have x= A

so, A = A sin (w(0)+ phi)

sin phi = 1

so, phi = pi/2 radians

so, x = 55 sin(wt + pi/2)

w = 2*pi(sqrt(k/m)) = 2*pi*sqrt(473/2.1) = 94.297

at t=0.31 s,

x= 55 sin (94.297*0.31 +pi/2) = -31.636 mm, i.e 31.636 mm to left

at t=1.07

x= 55 sin (94.297*1.07 +pi/2) = 51.34 mm,

t = 1.88

x= 55 sin (94.297*1.88 +pi/2) = 12.09 mm,

t = 2.47

x= 55 sin (94.297*2.47 +pi/2) = 49.86 mm,

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.