Simple Harmonic Motion -_- (1 Viewer)

[eskimoh]

i know this is an easy question but ive forgotten how to do it/cant do it for some reason!
helppppp please!

particle moves in SHM. at the extremities of the motin the absolute value of the acceleration is 1 cm/s2 and when the particle is 3 cm from the centre of the motion the speed is 2(squareroot)2 (ie. root 8) cm/s. find the period and amplitude

i got that
v^2 = n^2 (a^2 - x^2)
so 8 = n^2 (a^2 - 9)

and that acceleration = - n^2x
so 1 = -n^2x (at the extremity)

then i tried playing around with those two and i didnt get the answer-_-
thnx for ur help~

 

[eskimoh]

^ooh but the answers say:
T= 6pi
and the amplitude is 9 cm

 

[eskimoh]

Heya Moh! So here's what i got:

x: = -n^2 (x)

1 = - n^2 (3)

n^2 = 1/3

Then:

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

8^2 = 1/9 (a^2 - 9)

72 = a^2 - 9

a = sqrt (81)

a = 9

And:

T = 2 pi / n

T = 2 pi / (1/3)

T = 6 pi

ooh thnx but how did u know to put in 3 as x for the acceleration section.. cos it just said that acceleration was 1 at the extremities i think o_o
an d in ur working out.. u said n^2 = 1/3 but then u continued in the next bit and made n^2 = 1/9 and wher did th enegative from -n^2 go?

and where did the 72 come from o_o lol. after u squared 8 and made the right side a common denomiator of 9, where did the 9 go?

LOL sorry if i soudn rude! i just got confused in the working out but u got the right answers!

 

[eskimoh]

^lol oh ok
its pg 184 q 14 of the blue couchman txtbook!

 

[eskimoh]

LMAO its not hwk im studying ahahah and failing to do so as u can see!


hey sorry who is this again? hahaha i cant make out from ur accnt name! my bad!!