A particle moves along the x axis. It is initially at the position 0.210 m, movi

Question

A particle moves along the x axis. It is initially at the position 0.210 m, moving with velocity 0.200 m/s and acceleration -0.200 m/s^2. Suppose it moves with constant acceleration for 3.40 s. (a) Find the position of the particle after this time. M (b) Find its velocity at the end of this time interval. m/s We take the same particle and give it the same initial conditions as before. Instead of having a constant acceleration, it oscillates in simple harmonic motion for 3.40 s around the equilibrium position x = 0. (c) Find the angular frequency of the oscillation/s (d) Find the amplitude of the oscillation. Your response differs from the correct answer by more than 10%. Double check your calculations. m (e) Find its phase constant phi_0 if cosine is used for the equation of motion. Your response differs from the correct answer by more than 10%. Double check your calculations. rad (f) Find its position after it oscillates for 3.40 s. Your response differs from the correct answer by more than 100%. m (g) Find its velocity at the end of this 3.40 s time interval. differs from the correct answer by more than 100%. m/s

Explanation / Answer

d) Using law of conservation of energy

energy at the position x = 0.2 m = energy at the extreme position

(0.5*k*x^2) + (0.5*m*v^2) = (0.5*k*A^2)

k = m*w^2 =

0.5*m*w^2*x^2 + 0.5*m*v^2 = 0.5*m*w^2*A^2

m cancels

(0.5*0.953*0.21^2)+(0.5*0.2^2) = (0.5*0.953*A^2)

A = 0.293 m

------------------------------------------------------------

e) x = A*cos(phi_o)

cos(phi_o) = x/A = 0.21/0.293 = 0.717

phi_o = cos^(-1)(0.717) = 44.2 deg or 0.772 rad

f) x = A*cos(wt+phi_o)

x = 0.293*cos((0.976*3.4)+0.772)) = -0.171 m

g) v = w*sqrt(A^2-x^2) = 0.976*sqrt(0.293^2-0.171^2) = 0.232 m/sec

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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