anonymous
  • anonymous
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?
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
JamesJ
  • JamesJ
You mean its period is T?
anonymous
  • anonymous
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
  • anonymous
A2pie/T=A^2-X^2 its complex ... am i doing sm mistake because its becoming complex means i had done some mistakes anywhere

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

JamesJ
  • JamesJ
Please clarify. T is the period of the system?
anonymous
  • anonymous
yea
anonymous
  • anonymous
hey for pie^2 we can take it approx 10 rit
anonymous
  • anonymous
as its value is 3.14 so by squaring we can take approx 10??
JamesJ
  • JamesJ
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
  • anonymous
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
  • JamesJ
Look at the wording of the question and ask yourself what happens when t = 0.
anonymous
  • anonymous
wait a min they are asking time t rit or displacement??
JamesJ
  • JamesJ
Both I think. This is where writing the question in clear English is so important.
anonymous
  • anonymous
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
  • anonymous
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
  • JamesJ
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
  • anonymous
i got it i got it nw dont solve plz......
JamesJ
  • JamesJ
I'm solving from first principles. But I'll stop if you really want.
anonymous
  • anonymous
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
  • JamesJ
I respect that.
anonymous
  • anonymous
thanQ listen i m in confusion wid other i m nt able to make eaquation plz help me

Looking for something else?

Not the answer you are looking for? Search for more explanations.