24amplitude par equilibrium condition hogi
=> kA=mg =>A=mg/k
ab kuch solve hua???????
13 Answers
21isse clear nahi aata hai bhai!!! LOL... lagta che jeet gaya to tum gum mein duub gaye......
optn A : hor line : L max : mg/k above hor. line
optn A : hor line : L max : 2mg/k above hor. line
optn A : hor line : L + Mg/K max : mg/k above hor. line
optn A : hor line : L + Mg/K max : 2mg/k above hor. line
2421nahi gimme the ans.... acc to ur xplanation i guesss ans wud be A rite???
but A is not da Ans.
This que is frm tiit PART TEST 3
3well, initially, is the spring at equilibrium position or is it at its natural length ??
21THIS IS WAT DA SOLTN READS :
27 (c) z = (L + mg/k) - (mg/k)cos(ωt)
3ya dats wat...so at t=0, Z shud be equal to L not L + mg/K...dats y i told B....
bcoz its equilibrium pos is L + mg/k..thn a SHM of ampl mg/k....
21YAH! ME 2 THOT SAME, so do u xpect a que like dis 2 cum or it shud hav bn more specific rite!
3dunno for sure dude...the concept of da que isnt wrng but jus da grphs in the options...m not too sure...nwys, i got to finish sum mac...so u carry on...
21ha ha... k k... "thou mayst take thy leave" LOL.... macbeth's ghost haunts me