Question: A particle moves along the x axis. It is initially at the position 0 . 3 4 0 m , moving with velocity 0 .

A particle moves along the x axis. It is initially at the position 0.340 m, moving with velocity 0.220 m/s and acceleration -0.300 m/s2. Suppose it moves with constant acceleration for 5.30 s.
(a) Find the position of the particle after this time.
-2.7075
-2.71
m
(b) Find its velocity at the end of this time interval.
-1.37
-1.37
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 5.30 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.
0.939
0.939
/s
(d) Find the amplitude of the oscillation. Hint: use conservation of energy.
0.4129
0.413
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.
-0.603
-0.603
rad
(f) Find its position after it oscillates for 5.30 s.
-0.137
-0.137
m
(g) Find its velocity at the end of this 5.30 s time interval.
0.365
0.366
m/s

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