Question

Two identical springs of force constant 'k' are Joined one at the end of the other (in series). Find the effective force constant of the combination.

Answer 1

We know that if 2 springs are connected in series ( with spring stiffness say k for both) is given by:

1/K (eq)= 1/K + 1/K

Therefore, K(eq) = K/2

And formula for frequency of oscillation of a mass attached to a spring is given by:

f= 1/2 *pie sqrt ( k/m)

So substituting k= k/2 in the above formula gives the frequency of oscillation of mass m in case of series arrangement of springs.

Now, if the 2 springs are connected in parallel then K(eq) =K+K

Therefore, K(eq)=2K

Again substituting 2K in place of K in the frequency formula we get the value of frequency for parallel arrangement of springs.

For ratio, just the divide the frequency obtained in parallel to the frequency obtained in the series arrangement.( As mass of block is taken same for both arrangement).

Answer 2

Hii there,

● Answer-
k = (1/k1+1/k2)^-1

● Explaination-
When two springs are attached end to end, they are said to be in series combination.

Suppose k1 & k2 be the spring constants for two springs connected in series.

From Hooke's law-
For spring 1, F = k1x1
For spring 2, F = k2x2
Resultant F = kx

But total deformation (x) is -
x = x1 + x2
F/k = F/k1 + F/k2
1/k = 1/k1 + 1/k2

Therefore,
k = (k1^-1+k2^-1)^-1

Hope that is useful...