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Mechanics

Cool goIITian

Joined:	9 Oct 2007
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21 Dec 2007 10:46:05 IST

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433

Guys help me with this problem in IRodov

None

Refer problem 1.129

Wen 2 springs of spring constant K1 and K2 when connected together How does the resultant become

K1K2 / (K!+K2) plz help me!!!!!!!

ANd another doubt ..........

There is a spring connected to wall stiffness k

Wen i exert a force F to extend it x metres without imaprting any KE to it.

the equation for its motion is

F- kx = ma where a is some acceleration

OR That Forece F-Kx=0

DOES it Accelerate or NOT????????

I have doubt

i think it decelerates ... as wen u apply EnerGy conseravtion we find that

F=half kx ...............HOw do we find that deceleation using newtoons law????????

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Comments (7)

Rajat Khanduja

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21 Dec 2007 11:14:24 IST

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If you don't have change in kinetic energy, that means you have no change in velocity, then we can't have acc. 9 as you said we aren't imparting K.E.(i assume it to be about every dt time). Since no K.E. no acc.

If the force is not constant, then the kx also changes and it is possible not impart velocity to it.

Anand Hegde

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21 Dec 2007 15:42:58 IST

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for series connection of springs, use
1/K_eff = 1/K₁ + 1/K₂

Anand Hegde

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21 Dec 2007 15:43:42 IST

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For parallel connection use
K_eff = K₁ + K₂

Rajat Khanduja

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21 Dec 2007 15:44:59 IST

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@ anand.

But how do you get this formula.
Is there any other method.

Karthik M

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21 Dec 2007 15:51:21 IST

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For series connection, the elongations are different, the load is same,

So, F/K = F/K' + F/K''

Hence the result

Anand Hegde

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21 Dec 2007 15:51:48 IST

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2 springs are said to be connected in series if the total displacement of the hanging mass is equal to the sum of elongations in the springs.
X = x₁ + x₂
Each spring experiences the same tension F.
F = k₁x₁ which implies x₁ = F/k₁
F = k₂x₂ which implies x₂ = F/k₂

X = F/K_effF [1/k₁ + 1/k₂] = F/K_eff
1/k₁ + 1/k₂= 1/K_eff

Anand Hegde

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Joined: 12 Jun 2007

Posts: 1078

21 Dec 2007 15:52:51 IST

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