a particle starts SHM from the mean position. its amplitude is A and time is T.at what time when its speed is half of the maximum speed, its displacement is y is?

Question

a particle starts SHM from the mean position. its amplitude is A and time is T.at what time when its speed is half of the maximum speed, its      displacement is y is?

jamesj · Answer

You mean its period is T?

anonymous · Answer

here i did v^2=A^2-X^2
and here we know max velocity is Aw=vmax
w=2pie/T 
but nw i m unable to solve it further by keeping its value in the above formula

anonymous · Answer

A2pie/T=A^2-X^2
its complex ... am i doing sm mistake because its becoming complex means i had done some mistakes anywhere

jamesj · Answer

Please clarify.  T is the period of the system?

anonymous · Answer

yea

anonymous · Answer

hey for pie^2 we can take it approx 10 rit

anonymous · Answer

as its value is 3.14 so by squaring we can take approx 10??

jamesj · Answer

Ok then.  In which case the angular frequency $ \omega $ is given by
$$ \omega = \frac{2\pi}{T} $$
and the equation of motion of the system is
$$ x(t) = A \cos( \omega t + \phi) $$

Now, to solve the problem, first find $ \phi $ and then use this equation to find the value of $ t $ for which $ v(t) = \dot{x}(t)  $ is half the maximum velocity, which you'll also want to find.

anonymous · Answer

how can we find phi will u plz solve dis ques i think i need a solution here to see where i m lacking

jamesj · Answer

Look at the wording of the question and ask yourself what happens when t = 0.

anonymous · Answer

wait a min they are asking time t rit or displacement??

jamesj · Answer

Both I think.  This is where writing the question in clear English is so important.

anonymous · Answer

in my text they always write in a twisted form and i cant fight with education system alone to write it in good english :P

anonymous · Answer

hey i just rechecked the formula i used it was wrong the formula must be
v^2=w^2A^2-A^2x^2
my mistake i got my ans now :P

jamesj · Answer

To find phi: we're told that
"a particle starts SHM from the mean position" hence x(t=0) = 0.  But
$$ x(0) = A \cos(0 + \phi) $$
thus $ \cos \phi = 0 $ and therefore $ \phi = \pi/2 $.  

We can now write
$$ x(t) = A\cos(\omega t + \phi) = A\sin(\omega t) $$

anonymous · Answer

i got it i got it nw dont solve plz......

jamesj · Answer

I'm solving from first principles.  But I'll stop if you really want.

anonymous · Answer

yea plz actually if i see solution i dont use my mind dat is bad u cant help me in exam time dats y ...

jamesj · Answer

I respect that.

anonymous · Answer

thanQ listen i m in confusion wid other i m nt able to make eaquation plz help me

anonymous · Answer

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