|May1-11, 04:37 PM||#1|
Circular motion, oscillatory motion, SHM in springs
Points A,B,C,D, and E lie in a straight line. AB=BC=15 cm, CD=10 cm and DE=20 cm. A particle is moving with SHM so that A and E are the extreme positions of its motion. The period of the motion is 0.2s. Find the time the particle takes to get from B to D
i) if it is travelling towards D as it passes through B
ii) if it is travelling away from D as it passes through B.
w=2pi/T (w=angular speed, pi=3.14159, T=period)
v^2=w^2(a^2-x^2) (v=velocity, w=angular speed, a=amplitude, x=stretched distance)
My attemptFirst of all, the given answer is i) 0.0275s ii) 0.0942s
So, because A and E are the extreme points, with AE=60cm, therefore the a=30cm
Because period is 0.2s, 2*pi/T=angular velocity(w), so w=10pi=31.416
velocity at B: I used the SMH formula for velocity, v^2=w^2(a^2-x^2), x is the stretched distance, which is 0.15m from the center, so v^2=(10pi)^2*(0.3^2-0.15^2), v=8.162ms^-1
velocity at D: Same method applied, x=0.1m here. velocity at D=8.886ms^-1
From here on is just applied the distance formula with given distance, original velocity and final velocity. I think I messed up at this stage.
And finally deriving 0.029s, close but wrong
Thanks for the help
|May1-11, 04:54 PM||#2|
Couldn't you just use x = 0.3*sin(ωt) and evaluate t at x = -.15 and x = 0.1? Better yet, put the x equation into your graphing calculator and see what it is for all times.
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