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indr12365

Three springs of spring constant k are attached to a common mass 'm' and are equally inclined to each other.Time period of oscillation is:

{

Springs 120 apart.Each attached 2 fixed point at one end and 2 mass 'm' at their junction.

}

I've the ans as 2

2m/3k.

please detail in steps..

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aditi_g

c its simple...
just use energy concept
extension in each spring is x but along either one axis either x or y if we consider then for one spring it will be x and the other two it wud be x cos 60 coz they r inclined at an angle of 120
E=1/2k x^2 + 1/2 k (xcos60)^2+1/2 k (xcos60)^2+1/2 mv^2
we know dE/dt=0
u will get 3/2 kxv+ mva=0
from this u will get a=-3/2 kx
a=-w^2x
and T= 2pi/w
therefore T=2pi sqrt(2m/3k).
i hope it is clear...
if u have any doubt nudge me

sboosy

Assume it is displacd by x straight wrt to one spring...then correspondingly the extensions in the other two springs will be xcos(60) each

The force eqn (which is not much different,would be)

F = kx+

(k²x²cos²(60)+k²x²cos²(60)+2k²x²cos²(60)cos(120))

(Using the formula of vectors .a+b(vectors)(magnitude) =

(a²+b²+2abcos(alpha)

where alpha is the angle between the vectors a and b

Thus F = kx+kx/2 =3kx/2

T = 2

(m/k_eff) where k_eff in this case is 3k/2

Thus T = 2

(2m/3k)

indr12365

Both the methods described above take u 2 da correct answer,but certain things r wrong..

indr12365

Assume the spring system in da plane of screen
[x-y plane].
Now u r pushing the mass along -z axis[inwards].
Here da ques requires tat 3 springs b inclined equally 2 each other.
there4 at max displacement, da springs r inclined at 60 with z- axis.
Now da extentions produced on each spring is da same[ say 'x ']. Resolving da forces along x-y plane & +z axis,forces in x-y plane cancel out

abrambenny

Hey, the question says that the springs r equally inclined when in equilibrium... not after they have been displaced...

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