A particle moves along the x axis. It is initially at the position 0.200 m, movi

Question

A particle moves along the x axis. It is initially at the position 0.200 m, moving with velocity 0.140 m/s and acceleration -0.410 m/s2. Suppose it moves with constant acceleration for 5.50 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 5.50 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.

a) Find the angular frequency of the oscillation. Hint: in SHM, a is proportional to x.

b) Find the amplitude of the oscillation. Hint: use conservation of energy.

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.

d) Find its position after it oscillates for 5.50 s.

e) Find its velocity at the end of this 5.50 s time interval.

Explanation / Answer

Given that

The initial position of the particle is (x0) =0.200m

The initial velocity of the particle is(vo) =0.140m/s

The acceleration of the particle is (a) = -0.410 m/s2

The interval is given as(t) =5.50s

First we need to find the position of the particle, it is given by the formula

x =xo+vot+(1/2)at2

=0.200+(0.140)(5.50)+(0.5)(-0.410)(5.50)2 =0.200+0.77-6.201=-5.231m

The final velocity of the particle is given by (v) =vo+at =0.140+(-0.410)(5.50)=

0.140-2.255 =- 2.115m/s

a)

The angualr frequency is given by w =Sqrt(a/x) =Sqrt(0.410/5.231) =0.279rad/sec=0.28rad/s

b)

We know that the velocity of the simple harmonic oscillator is v =wSqrt(A2-x2)

Then the Amplitude is given by A =Sqrt((v/w)2+x2) =Sqrt(-2.115/0.28)2+(-5.231)2 =Sqrt(57.056+27.363)=9.188m

c)

The displacemnt of the particle is given by

x =Acos(wt+fi)

Then the phase angle fi =cos-1(x/A) =cos-1(-5.231/9.188) =55.296degrees

d)

The position of the particle after 5.50s is

x =Acos(wt+theta)

=(9.188)cos(0.28*5.50+55.296)=5.026m

e)

The final velocity of the particle after t=5.50s is

v =Awsin(wt+theta)

=(9.188)(0.28)sin(0.28*5.50+55.296) =2.153m/s

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