# M3 questionWatch

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#1
Hi
Points O,A,B lie on a straight line in that order. A particle P is moving on the line with SHM of period 4s, amplitude 0.5m and centre O. OA is 0.1m, OB is 0.3m. When t=0, P passes through B travelling in the direction OB. Calculate the time when P reaches A.

I said that the time for it to go from 0.3 above the equilibrium line back to it the line is 2s - time for it to go from O to B. I then added this to the time it takes to go from O to A. However, I dont get the right answer. :'(
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14 years ago
#2
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#3
(Original post by dvs)
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#4
(Original post by lgs98jonee)
ok well if you are not going to post solution unless i tell you the answer....the answer is 1.46s
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14 years ago
#5
(Original post by lgs98jonee)
I don't have one... I wanted to work until I find it.
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14 years ago
#6
(Original post by lgs98jonee)
ok well if you are not going to post solution unless i tell you the answer....the answer is 1.46s
Use x=Acos(wt). [A=0.5 & w=2pi/T=2pi/4=0.5pi]

Find t to reach point O from B:
0.3=0.5cos(0.5pi t)
t=0.59

Find t to reach A from O:
0.1=0.5cos(0.5pi t)
t=0.87

t=0.87+0.59=1.46
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#7
(Original post by dvs)
Use x=Acos(wt). [A=0.5 & w=2pi/T=2pi/4=0.5pi]

Find t to reach point O from B:
0.3=0.5cos(0.5pi t)
t=0.59

Find t to reach A from O:
0.1=0.5cos(0.5pi t)
t=0.87

t=0.87+0.59=1.46
So the question wanted to know the time from B down to A and not the time from B up to top and all the way back down to A? If that is the case then :'(
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#8
Ok...another question

A particle of mass 0.2kg is attached to an elastic string of modulus 15N and nat. length 1m to a point A of the smooth horizontal surface on which P rests. P receives an impulse of magnitude 0.5Ns in the direction AP. Show that while the string is taught, the motion of P is SHM and calculatre the period and amplitude of the motion.

Can not get the amplitude!!!!!!!!!!

T=0.726s
a=0.289m btw
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#9
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14 years ago
#10
(Original post by lgs98jonee)
Ok...another question

A particle of mass 0.2kg is attached to an elastic string of modulus 15N and nat. length 1m to a point A of the smooth horizontal surface on which P rests. P receives an impulse of magnitude 0.5Ns in the direction AP. Show that while the string is taught, the motion of P is SHM and calculatre the period and amplitude of the motion.

Can not get the amplitude!!!!!!!!!!

T=0.726s
a=0.289m btw
I'll try to do this without a sketch.
F = ma (a denoting acceleration here)
-T = ma
-(15)(x)/(1) = a/5
-(15)(5)x = a
a = -75x. As a=-w^2 x the motion is simple harmonic. Period = 2pi/w = 2pi/rt75 = 0.726s (2sf).
At x=0 it receives an impulse of 0.5Ns
As I = M(v-u).
The velocity at x=0 is given by: 0.5=0.2(v-0)
0.5/0.2 = v = 2.5
v^2 = w^2(a^2 - x^2) (a denoting amplitude here)
Substitution gives:
(2.5^2) = (75)(a^2 - 0)
Giving a= 0.289m

Sorry about the confusion between the 'a's but i can't type the second derivative of x on my keyboard.
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#11
(Original post by Gaz031)
I'll try to do this without a sketch.
F = ma (a denoting acceleration here)
-T = ma
-(15)(x)/(1) = a/5
-(15)(5)x = a
a = -75x. As a=-w^2 x the motion is simple harmonic. Period = 2pi/w = 2pi/rt75 = 0.726s (2sf).
At x=0 it receives an impulse of 0.5Ns
As I = M(v-u).
The velocity at x=0 is given by: 0.5=0.2(v-0)
0.5/0.2 = v = 2.5
v^2 = w^2(a^2 - x^2) (a denoting amplitude here)
Substitution gives:
(2.5^2) = (75)(a^2 - 0)
Giving a= 0.289m

Sorry about the confusion between the 'a's but i can't type the second derivative of x on my keyboard.
Thanks and ugh also...I decided to use the value for T rather than w to find the amplitude
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