16-03-09 Simple harmonic Motion

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 ρ, while that of water is 1

the spring constant is k

Is this a SHM. Find the time period.

11 Answers

106
Asish Mahapatra ·

i think it is SHM ...

equilibrium posn is when spring ext. is y = (ρga3 - ρwga3/2)/k

if we displace the body downward by x from this equilibrium position...

k(y+x) - mg - ρwga2(a/2 + x) = ma...

it becomes
a = (k+ρwg)x/ρa3

so ω2 = (k+ρwg)/ρa3

so T = 2Ï€√ρa3/(k+ρwg)

62
Lokesh Verma ·

take the case when there was no water.. then wud ρw=0 wud give the shm of a spring?

106
Asish Mahapatra ·

i din get u ...

yes then also it wud be SHM ..

with T = 2Ï€√ρa3/k = 2Ï€√m/k

just put ρw = 0

21
tapanmast Vora ·

Wud this not be damped oscillation type?

62
Lokesh Verma ·

why damped tapan?

106
Asish Mahapatra ·

well if u consider all those resistive forces such as air resistance... viscosity ..... then it will be damped... otherwise .. i dont think it will

13
deepanshu001 agarwal ·

2pi√m/k+a^2gp d ans

21
tapanmast Vora ·

is this a case wer : CUbe goesinto da water n cums out, this cylce continues.....

then dampness i thot wud b cozd by WATER

21
tapanmast Vora ·

in POST #4

Asis : then its a simple SHM,

has 2 b T = 2pi√mg/k or is 2pi root (m/k)

13
deepanshu001 agarwal ·

jus put k effectiv in d generalized formula

btw nybody in favour of my ans ...?

106
Asish Mahapatra ·

tapan.. thx for pointing out.. i ve corrected the original anser ... now i think u can get wat u wanted..

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