motion problem (1 Viewer)

[followme]

I'm confused with solving SHM questions, please help.. thanks.
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A particle moving with SHM starts from rest at a distance 6m from the centre of oscillation. If the period is 4pi seconds:
a) find the time taken to move to a point <?xml:namespace prefix = st1 ns = "urn:schemas-microsoft-com

6m</st1:chmetcnv> from the centre of oscillation. If the period is 4pi seconds: <O></O>
<font face=" /><st1:chmetcnv UnitName="m" SourceValue="3" HasSpace="False" Negative="False" NumberType="1" TCSC="0" w:st="on">3m</st1:chmetcnv> from the origin<O></O>
<FONT face=Arial><FONT size=3>b) find the velocity and acceleration at this point<O></O>
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<FONT face=Arial><FONT size=3>period = 4π<O></O>
<FONT face=Arial><FONT size=3>4π=2π/n<O></O>
<FONT face=Arial><FONT size=3>n=1/2<O></O>
<FONT face=Arial><FONT size=3>v<SUP>2</SUP>=n<SUP>2</SUP>(A<SUP>2</SUP>-x<SUP>2</SUP>) , 0=1/4(A<SUP>2</SUP>-36) , so A=6<O></O>
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<FONT face=Arial><FONT size=3>a) for SHM, x(t) can be : x(t)=Asin(nt+α) or x(t)=Acos (nt+β)<O></O>
<FONT face=Arial><FONT size=3>since t=0 x=6 <O></O>
<FONT face=Arial><FONT size=3>so it is 6=6sin (0/2+α) , α= π/2 or 6=6cos (0/2+β), β=0<O></O>
<FONT face=Arial><FONT size=3>Therefore the equations to first part are 3=6sin(t/2+π/2) or 3=6cos (t/2) however the sine one gives t= -2π/3 and the cosine one gives t=2π/3. They differ in sign, so I’m not sure which one to use. On other occasion, the cosine one is –tive and the sine one is +tive. <O></O>
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<FONT face=Arial size=3>b) v<SUP>2</SUP>=n<SUP>2</SUP>(A<SUP>2</SUP>-x<SUP>2</SUP>) =27/4 , so v= ±√27/2 However the book's answer is positive √27/2. How to determine the sign?

 

[pLuvia]

Can you please re type the question it's really hard to read

 

[insert-username]

Therefore the equations to first part are 3=6sin(t/2+π/2) or 3=6cos (t/2) however the sine one gives t= -2π/3 and the cosine one gives t=2π/3. They differ in sign, so I’m not sure which one to use. On other occasion, the cosine one is –tive and the sine one is +tive.

Time must be a positive value. There is no such thing as negative time, so you therefore use the positive value (i.e. take the absolute value of the result).

b) v2=n2(A2-x2) =27/4 , so v= ±√27/2 However the book's answer is positive √27/2. How to determine the sign?

I think that in this case, velocity could be positive or negative depending on the direction of the particle, so you use the absolute value.


I_F

 

[shsshs]

u say 6m from the centre.
this could be +6 or -6
so cldnt ur amplitude be -6?

 

[SoulSearcher]

Amplitude is a distance, therefore you should take the absolute value, in this case 6m