Question

A particle moves along the x axis. It is initially at the position 0.230 m, moving with velocity 0.200 m/s and acceleration -0.420 m/s2. Suppose it moves with constant acceleration for 5.30 s.

(a) Find the position of the particle after this time. m

(b) Find its velocity at the end of this time interval. 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. /s

(d) Find the amplitude of the oscillation. Hint: use conservation of energy. m

(e) Find its phase constant ϕ0 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 5.30 s. m

(g) Find its velocity at the end of this 5.30 s time interval. m/s

Answer 1

a) x = xo + u*t + 0.5*a*t^2

= 0.23 + 0.2*5.3 + 0.5*(-0.42)*5.3^2

= -4.61 m

b) v = u + a*t

= 0.2 + (-0.42)*5.3

= -2.03 m/s

c) F = k*x

m*a = k*x

a/x = k/m

we know, angular frequency, w = sqrt(k/m)

= sqrt(a/x)

= sqrt(0.42/0.23)

= 1.35 rad/s

d) F = -k*x
m*a = -k*x

k = -m*a/x

= -m*(-0.42)/0.23

= 1.83*m

(1/2)*k*A^2 = (1/2)*k*x^2 + (1/2)*m*v^2

(1/2)*1.83*m*A^2 = (1/2)*1.83*m*0.23^2 + (1/2)*m*0.2^2

(1/2)*1.83*A^2 = (1/2)*1.83*0.23^2 + (1/2)*0.2^2

A = 0.27 m

e) x = A*cos(w*t + phi)

at t = 0

x = A*cos(phi)

==> phi = cos^-1(x/A)

= cos^-1(0.23/0.27)

= 0.55 rad