SHM Fitzpatrick (1 Viewer)

[edmund gosse]

Exercise 25(c), p.113, question 7

Solve the differential equation d^2x/dt^2 + 16x =0 subject to the conditions x=3 and dx/dt = 16 when t=0. Find the maximum displacement and the maximum speed if x metres is the displacement of a particle moving in a straight line at time t seconds.

How is this done? How typical is this question of HSC SHM questions?

 

[Timothy.Siu]

Exercise 25(c), p.113, question 7

Solve the differential equation d^2x/dt^2 + 16x =0 subject to the conditions x=3 and dx/dt = 16 when t=0. Find the maximum displacement and the maximum speed if x metres is the displacement of a particle moving in a straight line at time t seconds.

How is this done? How typical is this question of HSC SHM questions?

it just means that a=-16x

 

[edmund gosse]

Yeah, that's what I thought, but...

 

[Timothy.Siu]

vdv/dx=-16x
0.5v^2=-8x^2+C

x=3, v=16
128=-72+C
C=200
v^2=-16x^2+400
max speed is when a=0, a=0 when x=0
so v^2=400 max speed is 20
max displacement is when v=0
16x^2=400
x=+-20/4

 

[FFC]

Alternatively,

use equation to find "n" - as for SHM, accel = -n^2x

n = 4

x = Acos(nt+k) --> @ t=0, x=3 THEREFORE 3 = A cos(k) --------- (1)
dx/dt = -nAsin(nt+k) --> @ t=0, v=16 THEREFORE 16 = -4Asin(k) ----------- (2)

By division of 1 and 2

k = arctan(-4/3)

Substitution into either of 1 or 2 gives A = 5

Therefore, Max Displacement is 5m (as must be the positive value), and max speed = abs(5*4) = 20