A cube of mass of side a is dipped in water through a spring.
At equilibrium , half the cube isimmersed in water
The density of the cube is Ï, while that of water is 1
the spring constant is k
Is this a SHM. Find the time period.
106i think it is SHM ...
equilibrium posn is when spring ext. is y = (Ïga3 - Ïwga3/2)/k
if we displace the body downward by x from this equilibrium position...
k(y+x) - mg - Ïwga2(a/2 + x) = ma...
it becomes
a = (k+Ïwg)x/Ïa3
so ω2 = (k+Ïwg)/Ïa3
so T = 2Ï€√Ïa3/(k+Ïwg)
62take the case when there was no water.. then wud Ïw=0 wud give the shm of a spring?
106i din get u ...
yes then also it wud be SHM ..
with T = 2Ï€√Ïa3/k = 2Ï€√m/k
just put Ïw = 0
106well if u consider all those resistive forces such as air resistance... viscosity ..... then it will be damped... otherwise .. i dont think it will
21is this a case wer : CUbe goesinto da water n cums out, this cylce continues.....
then dampness i thot wud b cozd by WATER
21in POST #4
Asis : then its a simple SHM,
has 2 b T = 2pi√mg/k or is 2pi root (m/k)
13jus put k effectiv in d generalized formula
btw nybody in favour of my ans ...?
106tapan.. thx for pointing out.. i ve corrected the original anser ... now i think u can get wat u wanted..