Consider a parcel starting when t = 0 at a vertical displacement s of 5 m from i

Question

Consider a parcel starting when t = 0 at a vertical displacement s of 5 m from its altitude z_0 before being displaced, and vertical velocity component w_0 = 2 cm/sec. For a solution s(t) to equation (1) to describe its motion, the solution should satisfy s(0) = 5m, ds/dt(0) = 2 cm/sec (3) Assume that N^2 is a non-negative constant and define s(t) = A cos(N t) + B sin(Nt) Find the values A and B that give a solution that satisfies the initial conditions (4). The value of N isn't being specified so you may expect that your answer may be an expression involving N. Find the maximum upward displacement of the parcel. Notice that this answer definitely depends on the value of N. Give a physical explanation for why increasing the value of N should have the effect of changing the maximum displacement in (b).

Explanation / Answer

given equation

s(t) = Acos(Nt) + Bsin(Nt)

boundary conditions

s(0) = 5 m

ds/dt (0) = 2 cm/s

a. so s(0) = A = 5 m

ds/dt = -ANsin(Nt) + BNcos(Nt)

ds/dt (0) = BN = 2 cm/s

so, B = 2 cm/ N

hence equation becomes

s = (5 m) cos(Nt) + (2 cm/ N) sin(Nt)

b. s(max) is when ds/dt = 0 [ condition of maxima minima]

so, -ANsin(Nt) + BNcos(Nt) = 0

Asin(Nt) = Bcos(Nt)

500sin(Nt) = 2cos(Nt)/N

250sin(Nt) = cos(Nt)/N

tan(Nt) = 1/250N

Nt = arctan(1/250N)

t = (arctan(1/250N))/N

so s max = 5m *cos(arctan(1/250N)) + (2cm/N)sin(arctan(1/250N))

c. now, when N increases, arctan(1/250N) decreases, cos ot this value increases and hence s max increases, (also because the sin term is multipiled by 2 cm/ N which is small, and gets smaller when N increases)

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