Questions from 1986 HSC (1 Viewer)

[currysauce]

Hello i need help please.

First is SHM

a) show that x=acos(2t+b) is poss. equation... of a= -4x DID that

b) the particle is observed at t=0 to have a velocity of 2m/s and a displacement from the origin of 4 m. Show that the amp. is root(17)

can't do that or this

c) determine the max velocity of the particle.

 

[Pace_T]

b)

x = acos(nt+ b)
when t = 0, x = 4
.'. 4 = acos(b) (equation 1)
when t = 0, v = 2
.'. 2 = -ansin(b)

hmm, now im assuming n = 2 (from a))
.'. 2 = -2asin(b)
.'. -1=asin(b) (equation 2)
equation 2 divided by equation 1
.'. -1/4 = tan(b)
b = - tan^-1(1/4)

.'. x = acos(2t -tan^-1(1/4))
when t =0, x = 4
.'. 4 = acos(-tan^-1(1/4))
use the triangle, if tan^-1(1/4) = b, .'. b= cos^-1((4/root17))
.'. 4 = acos(-cos^-1(4/root17))
4 = -a(4/root17)
a = -root17
a= root17
disregard the negative as it is amplitude
i hope that's right, its confusing, i know, sorry lol

 

[Pace_T]

c) determine the max velocity of the particle.

max velocity when x = 0

.'. v^2 = n^2(a^2-x^2)

.'. V^2 = n^2(17)

V^2 = 4(17)
v = root(68)
=8.2m/s
i think that's right, not too sure though

 

[shafqat]

the answer to part c is correct, but our teacher always advised us not to assume that formula for v^2, but prove it from d/dx(v^/2) = -n^2x
alternatively differentiate the expression for x with respect to t to get velocity, then the max value is the max value of that trig curve