There are two bodies P and Q which are individually connected to 2 separate massless springs ,having force constant k1 and k2 respectively.If two bodies oscillate vertically having their max. velocities equal.the ratio of amplitudes Of P and Q is.....

Question

mani_jha · Answer

The maximum velocity is given by
$$v=Aw$$
w1=$$\sqrt{k1/m1}$$
Find w2. Then
A1w1=A2w2.
does this give u the answer?

anonymous · Answer

yup you pretty well know what you are talking,good approach.

mani_jha · Answer

thanks bro. check this question. http://openstudy.com/users/shankvee%5C#/updates/4f2e274ae4b0571e9cbabfca Do it yourself, before seeing the solution. Use the method I told u

anonymous · Answer

good one mate..

mani_jha · Answer

Dont see the answer until you do it! That;s one bad habit you've got

anonymous · Answer

hey i did that sum by my self

anonymous · Answer

i have the misfortune of doing the exact sum in my institute

anonymous · Answer

the maximum velocity of body which is connected to spring is a multiplication of amplitude and angular frequency of that body:
now let shm equation is$$x=a sin(wt+phi)$$ then velocity=$$aw cos (wt+phi)$$ the maximum value of velocity will be a*w;where w is angular frequency and a is amplitude.
let the max velocity of first block is v and other is u then their ratio is :$$v/u =a1*w1/a2*w2$$ given that v/u=1 hence solve it and find a1/a2 i.e equal to w2/w1=root of k2m1/k1m2.where m1 and m2 is the mass of two bodies respectively