An object undergoes simple harmonic motion in two perpendicular directions. The

Question

An object undergoes simple harmonic motion in two perpendicular directions. The object's position as a function of time is given as: Express the two dimensional harmonic oscillator's distance from the origin and select it from the answers below. Be sure to simplify your expression. Next, find the expression for the object's velocity in terms of the variables used in the position equation. Be sure to the hat decoration over any unit vectors you use. It takes the object 0.650 s to complete one oscillation. What is the object's speed if its distance from origin (radius of the circle) is 0.250 m?

Explanation / Answer

the first one is a bit tricky. The way you want to visualize it is as a 2D vector. The vector is r. x component of r is Lcos(wt) and the y component is Lsin(wt). we want magnitude. magnitude of vector is:

sq_root((x-comp)^2 + (y-comp)^2)

i.e. the answer to question 1 appears to be D. L*sq_root(cos^2 + sin^2). However (cos^2 + sin^2) = 1. So the answer is C, just L

2.

the velocity is same as the derivative of r. The x-cop is -L*w*sin(wt) and y-component is L*w* cos(wt). Write in the appropriate format. If you want ro wirte it in terms of unit vectores, just add i_hat to the x and y components.

3.we have radius = 0.250m. time to complete one oscillation = 0.650s. angles in 1 oscillation = 2pi radians.

So angular speed (w) = 2pi / 0.650 = 9.666radians per second

we have v = w*r = 9.666* 0.250 = 2.42 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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