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1/2 mv^2 = 1/2 kx^2
Why did you close this question?
for maximum compression..whole kinetic energy is transferred to spring energy...that helps ?
and in such case,,effective mass = m1 m2 / (m1 + m2) am i helping ?
i mean are you getting ?
wat is effective mass
2 masses are connected to a spring on 2 ends..say m1 and m2.. then the effective mass is m1 m2/ (m1 + m2)
Wait....I mean....I did nt get...wat is effective mass shuld nt it be m1 + m2
effective mass or can be also called reduced mass of the system.. never heard of it ?
So.... 1/2 k x^2 = 1/2 effective mass * v^2
well reduced mass is no big deal..has a simple derivation.. you wanna know ?
there isnt much need for the derivation for you to know since you're JEE aspirant..but still i can tell if you want..
Yes..i am interested in it
kk..newton's 3rd law states that F1 = -F2 right ? both taken in vector form.. F1 will be one acting on 1st body,,F2 on second..
all expressions are in vector form below : m1 a1 = - m2 a2 =>a1 = (-m2/m1) a2 now relative acceleration between the bodies a = a2 - a1 = a2 (1 + m2/m1) = m2 a2 ( m1 + m2)/(m1 m2) following ?
tell me if not getting anything ? we are not done yet with the thing..
Yup....i can follow u..)
Tell me a Scenario where we will use Effective Mass
now..m1m2 / (m1+ m2) = M (reduced or effective mass) so we have F2/M = a or F2 = Ma so what do you conclude ?
see,,any 2 body system can be converted to 1 body system with this effective mass... force acting between them will also act between this one body system..
Nw..i get it...)...thxx....)
hmm,,whats the conclusion you can draw then?