A particle moves along the x axis. It is initially at the position 0.330 m, movi

Question

A particle moves along the x axis. It is initially at the position 0.330 m, moving with velocity 0.140 m/s and acceleration -0.330 m/s*. Suppose it moves with constant acceleration for 3.20 s. (a) Find the position of the particle after this time -0.9116 (b) Find its velocity at the end of this time interval. -0.916 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 3.20 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. (d) Find the amplitude of the oscillation. Hint: use conservation of energy. (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 3.20s (9) Find its velocity at the end of this 3.20 s time interval m/s

Explanation / Answer

a) and b) are motion with constant linear acceleration

a)
x(t) = x(o) + v(o)×t + a×t²/2
x(3.2) = (0.33 m) + (0.14 m/s)×(3.2 s) + (-0.33 m/s²)×(3.2 s)²/2
x(3.2) = -0.9116 m

b)
v(t) = v(o) + a×t
v(3.2) = (0.14 m/s) + (-0.33 m/s²)×(3.2 s)
v(3.2) = -0.916 m/s

c) through g) are simple harmonic motion (SHM)

The equation for position of SHM as a function of time is
x(t) = A cos(t + )

The equation for velocity of SHM as a function of time is:
v(t) = dx/dt = d(A cos(t + ))/dt
v(t) = -A sin(t + )

The equation for acceleration of SHM as a function of time is:
a(t) = dv/dt = d(-A sin(t + ))/dt
a(t) = -A ² cos(t + )

where
A = amplitude
= angular velocity (or angular frequency)
= angular phase shift

Nothing is given about position, velocity or acceleration AT A SPECIFIC TIME, so you can choose the time to be t=0 when the particle is moving through x=0.33.
At that point you know that:
x(o) = A cos(×0 + ) = 0.33 m
v(o) = -A sin(×0 + ) = 0.14 m/s
a(o) = -A ² cos(×0 + ) = -0.33 m/s²
which simplifies to 3 equations with 3 unknowns:
A cos() = 0.33 m
A sin() = -0.14 m/s
A ² cos() = 0.33 m/s²

Solving this system of equations, you will find:
= 1 rad/s <- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - answer c
A = 0.358 m <- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - answer d
= -0.4012 rad <- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - answer e
(contact me if you don't know how to solve this system)

Your equations now become:
position:
x(t) = 0.358 × cos(1×t - 0.4012)

velocity:
v(t) = -0.358 × 1 × sin(1×t – 0.4012)
v(t) = -0.358*sin(t - 0.4012)

acceleration
a(t) = -0.358 × 1² × cos(1×t - 0.4012)
a(t) = -0.358*cos(t - 0.4012)

Filling in t = 3.2 into the equation for position, you get:
x(3.2) = 0.358 × cos(3.2 - 0.4012)
x(3.2) = -0.337 m <- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - answer f

Filling in t=3.3 into the equation for velocity, you get:
v(3.2) = -0.358*sin(3.2 - 0.4012)
v(3.2) = -0.1203 m/s <- - - - - - - - - - - - - - - - - - - - - - - - - - - - answer g

Hope it helps....!!!

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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