A particle moves along the x axis. It is initially at the position 0.140 m, movi

Question

A particle moves along the x axis. It is initially at the position 0.140 m, moving with velocity 0.240 m/s and acceleration -0.270 m/s2. Suppose it moves with constant acceleration for 3.10 s. (a) Find the position of the particle after this time. Incorrect: Your answer is incorrect. . Your response differs from the correct answer by more than 100%. m (b) Find its velocity at the end of this time interval. Incorrect: Your answer is incorrect. . Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. 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 3.10 s around the equilibrium position x = 0. Hint: the following problems are very sensitive to rounding, and you should keep all digits in your calculator. (c) Find the angular frequency of the oscillation. Hint: in SHM, a is proportional to x. /s (d) Find the amplitude of the oscillation. Hint: use conservation of energy. m (e) Find its phase constant 0 if cosine is used for the equation of motion. Hint: when taking the inverse of a trig function, there are always two angles but your calculator will tell you only one and you must decide which of the two angles you need. rad (f) Find its position after it oscillates for 3.10 s. m (g) Find its velocity at the end of this 3.10 s time interval. m/s

Explanation / Answer

given data

position =0.140m

velocity = 0.240m/s

acceleration a=-0.270m/s^2

t=3.10s

fom kinematic relations

(a) x = Xo + Vo*t + ½at²
x = 0.140 + 0.240*3.10 - 0.5*0.270*(3.10*3.10)= -41335 m

b) v = Vo + at = 0.240- 0.240*3.10= -0.504 m/s

If we model the motion with
x(t) = Asin(t + ), then
v(t) = Acos(t + ) and
a(t) = -A²sin(t + )

Given x(0) = 0.140 m and v(0) = 0.240m/s and a(0) = -0.270m/s²:

(c) x(0) / a(0) = 0.140m / -0.270m/s² = Asin(t + ) / -A²sin(t + )
-0.51851 s² = -1/²
= (1 / 0.51851) =1.388 rad/s

(d) TME is constant.
At t = 0, TME = ½kx² + ½mv² = ½k(0.140)² + ½m(0.240)²
at x = 0, TME = ½m(Vmax)² = ½m(A)² = ½mA²²
and at x = A, TME = ½kA²
So ½k(0.140)² + ½m(0.240)² = ½kA² divide by mass m
(k/m)(0.140)² + (0.240)² = (k/m)A² but k/m = ² = 1.926544, so
(1.926544)(0.140)² + (0.240)² = (1.926544)A²
solves to
A = 0.22248 m

(e) Well, I had used sine, but one could write
x = Acos(t + ), and at t = 0,
0.140 m = 0.22248m * cos(0 + )
cos = 0.62927
= ± 0.8901 rads
Do we need "+" or "-"?
Now v(t) = -Asin(t ± ), and when t = 0,
v(0) = -Asin(±)
But we know that v(0) < 0, so sin(±) > 0 which means that we need the "+" and not the "-" value.
x(t) = 0.22248m * cos(1.388*t + 0.8901)

(f) x(3.10) = 0.22248m * cos(1.388*t + 0.8901) = -0.1028 m

(g) v(3.10) = -0.22248* 1.388 * sin(1.388*t + 0.8901) = -0.2738 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.