Simple Harmonic Motion (1 Viewer)

[Kutay]

Hey i was wondering if anyone could show me how to do these questions...

1. "A Particle moves in S.H.M. with n = 2 . if its rest position is distant 4m from the origin, find its velocity as it passes through the origin"

2. "A point moving with S.H.M. has a velocity of 4m/s when passing through the centre of its path and its period is PI seconds. What is it velocity when it is 1m from the position in which its velocity is zero.

3. "The rise and fall of the tide appromiate smh. suppose the interval between successive high tides in a certain harbour is 12hours 30minutes and that in the entrance the depth of water at high tide is 10m and at low tide 6m. if on a certain day low tide occurs at 6.am at what time will the depth of water in the harbour entrance first reach 9m?"

Thank you

 

[name_disclosed]

Hi
For 1. i think this is how you do it
aceleration=-n^2x = -4x
therefore, vdv/dx = -4x
fvdv = -4fxdx
v^2/2 = -2x^2 + c
v=0, x = 4
therefore c = 32
factorising gives v^2 = 4(16-x^2)
hence, when x = 0, v^2 = 64, therefore v = 8 when x is at the origin

for 2. period = Pi = 2Pi/n therefore, n=2
from general equation: v^2 = n^2 (a^2-x^2) we get v^2 = 4(a^2-x^2)
but when x=0, v=4, so 16=4(a^2), therefore, a = 2
hence, the euqation becomes, v^2=4(4-x^2)
when its 1 m from v=0, its also 1 m from mean pstion, so v^2=4(4-1) = 12
so v = root12

for3. Peroid is 12.5 hours, so n = PI0.16
also, high tide and low tide are 10m and 6m respectively, so if we take 8m as the mean postion, the amplitude is 2m.
also, let 6am be T=0
Using x=acos(nt+alpha)
we get x = 2cos(PI0.16 t + alpha)
but taking t=0, and x = -2
there -1=cosalpha
so alpha = PI

therefore, x = 2cos(0.16PIt+ PI)
but when its 9m, x=1
so 1=2cos(PI0.16t+180)
hence, 5/2PI = PI0.16t + PI
so t = 9.375 hours from when t=0
therefore 6+9.375 = 3 :22 pm

im not sure about the last one though, i think its right

 

[dawso]

just have a l;ook at the bigger picture in ur answer there....

if its low at 6am, next low will be at 6:30pm......

so it wont be first at 8m high at 3:22, cause high will hav alreayd occured at 12:15pm....

i havent checked but ur answer may well be right but its not the first time which is what they want,

i agree wit answers for first 2, for the third i got 9:51 am

 

[name_disclosed]

How did you get 9:21?
Your logic makes sense and my 3:22 is certainly incorrect
But i couldnt get a value other than 3:22
Did you use asin(nt+alpha) instead of acos(nt+alpha)?

 

[dawso]

cant find the sheet i roughed it out on this morning so now ill just try 2 wing it...


T = 25/2

therefore n = 4PI / 25

(here is where i went wrong before)

a = 2m (i think i had 3 for some reason)

so x = 2 cos (4PI/25 t)

and its when x = 1

so 4PI/25 t = PI/3

so t = 25/12 = 2 hours 5 minutes

therefore its 9 metres high at 8:05 am (this doesnt sound rite 2 me and ive prob made a heap more mathematical fuk ups as i always do in 3unit so edit me away guys...)

 

[frootloop]

Actually what the last post is exactly what I got.

8:05am.

I'm not the greatest at 3U I'll admit but I agree on that last answer...

BTW I also agree with the other two answers as well (for q 1 and 2)