hi guys

     pls tell me this

          2 blocks of mass  m are connected by a string of spring cnst k---- they r displaced slightly in opp dirns and released. find the time period.

          now cant we directly say that =k/2m ? the answer given is

=2k/m  --- plz explain my mistake and wen can i use this methodof simply adding up the masses?

          i am not able to solve simple hcv probs- of shm --wat shd i do? is my basic mechanics wrong?

The only way to find  is to give small displacement dx to the body abd find the force acting on it. This will give force constant and from that you get . Please do not get into this buisness of getting to answers directly because what seems direct in one case may change in other. In this problem suppose total streching of dx is given to the string and left. In an inertial frame (center of mass frame is one such frame) each body is diplaced by dx/2 and force on each body is Kdx. So force constant for each body

 k' = F/(dx/2) = 2k

 = sqrt(2k/m)

dude....wen 2 or more masses r there u hv to introduce a concept of reduced mass....denoted by .........

   given by  = product of all the masses/ sum of all the masses....

 so in this case it is...... = m²/2m.....

   which gives u.... = m / 2

   hence  = 2k/m......

    if u like it thn plzzzzz rate me...!!!!!

Omi here also we what you are doing is applying a result directly without deriving it.

In my derivation if i had taken m1 and m2 instead of m and m i would have got omega = k(m1+m2)/(m1m2). From here we take out a direct result that when ever you have two masses m1 and m2 attached with spring of spring constant k assume that one is fixed and other has a mass of m1m2/(m1+m2) (called reduced mass) and find omega or time period directly.

And this thing should be absolutly clear to you that what i have done is derivation and what you are doing is just applying a result. It is OK to do so if you know the derivation and you are doing so just to save the time. But it is criminal to do so if you don't know how that result come. Because when it comes to applying same result in some other conditions you will have to make a decision whether you can apply it there are not. And you can do that only when you have an indepth understanding of result.