A particle moves along the x axis. It is initially at the position 0.130 m, movi

Question

A particle moves along the x axis. It is initially at the position 0.130 m, moving with velocity 0.080 m/s and acceleration -0.330 m/s2. Suppose it moves with constant acceleration for 3.60 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 3.60 s around the equilibrium position x = 0. (a) Find the angular frequency of the oscillation. Hint: in SHM, a is proportional to x. /s (b) Find the amplitude of the oscillation. Hint: use conservation of energy. m (c) 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 (d) Find its position after it oscillates for 3.60 s. m (e) Find its velocity at the end of this 3.60 s time interval. m/s

Explanation / Answer

At that point you know that:
x(o) = A cos(×0 + ) = 0.13 m
v(o) = -A sin(×0 + ) = 0.080 m/s
a(o) = -A ² cos(×0 + ) = -0.330 m/s²
which simplifies to 3 equations with 3 unknowns:
A cos() = 0.13 m
A sin() = -0.080 m/s
A ² cos() = 0.330 m/s²

Solving this system of equations, you will find:
a) = 1.5932 rad/s    0.79772
b) A = 0.145 m           0.44656
c) = -0.368 rad         -0.73923

Your equations now become:
position:
x(t) = 0.145 × cos(1.5932×t - 0.368)

velocity:
v(t) = -0.145 × 1.5932 × sin(1.5932×t - 0.368)
v(t) = -0.2310 sin(1.5932×t - 0.368)

acceleration
a(t) = -0.145 × 1.5932² × cos(1.5932×t - 0.368)
a(t) = -0.3680 cos(1.5932×t - 0.368)

d) Filling in t=3.6 into the equation for position, you get:
x(3.3) = 0.145 × cos(1.5932×3.6 - 0.368)
x(3.3) = -0.144 m

e)

Filling in t=3.6 into the equation for velocity, you get:
v(3.3) = -0.2310 sin(1.5932×3.6 - 0.368)

v(3.3) = 0.0216 m/s

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