Consider two massless springs connected in series. Spring 1 has a spring constant k1, and spring 2 has a spring constant k2. A constant force of magnitude F is being applied to the right. When the two springs are connected in this way, they form a system equivalent to a single spring of spring constant k.

Part A

What is the effective spring constant k of the two-spring system? Express the effective spring constant in terms of k1 and k2.

Part B

Now consider three springs set up in series as shown. (Figure 2) The spring constants are k1, k2, and k3, and the force acting to the right again has magnitude F. Find the spring constant k′ of the three-spring system. Express your answer in terms of k1, k2, and k3.

Question

Mckinney

Physics

11 January, 07:27

Consider two massless springs connected in series. Spring 1 has a spring constant k1, and spring 2 has a spring constant k2. A constant force of magnitude F is being applied to the right. When the two springs are connected in this way, they form a system equivalent to a single spring of spring constant k.

Part A

What is the effective spring constant k of the two-spring system? Express the effective spring constant in terms of k1 and k2.

Part B

Now consider three springs set up in series as shown. (Figure 2) The spring constants are k1, k2, and k3, and the force acting to the right again has magnitude F. Find the spring constant k′ of the three-spring system. Express your answer in terms of k1, k2, and k3.

+2

Answers (1)

Know the Answer?

Cordell Yoder · Answer 1

a) K = k1k2/k2+k1

b) k1k2k3 / (k2k3+k1k3+k1k2)

Explanation:

Hooke's law states that the extension of an elastic material is directly proportional to the applied force provided the elastic limit is not exceeded. Mathematically;

F = ke where;

F is the applied force

k is the elastic constant

e is the extension

If we consider 2 springs 1 an 2 with spring constant k1 and k2 connected in series to each other, their respective spring constant according to hooke's law will be expressed as;

k1 = F/e1 and k2 = F/e2 where F is a constant force.

e1 = F/k1 and e2 = F/k2.

The equivalent extension e = F/K

Since the two spring are in series, the effective spring constant K of the two-spring system is expressed as follows;

Since the total extension of the string

e = e1+e2

F/K = F/k1+F/k2

F (1/K) = F (1/k1) + F (1/k2)

1/K = 1/k1+1/k2

1/K = (k2+k1) / k1k2

Reciprocating both sides gives

K = k1k2/k2+k1

b) Similarly if there are 3 springs connected in series to each other with spring constant k1, k2 and k3, their individual extension will be expressed as;

e1 = F/k1

e2 = F/k2

e3 = F/ke

Their equivalent extension e in series will be expressed as e = F/K

Writing their equivalent extension in terms of their individual extension will give;

e = e1+e2+e3

F/K = F/k1+F/k2+F/k3

1/K = 1/k1+1/k2+1/k3

1/K = (k2k3+k1k3+k1k2) / k1k2k3

Taking the reciprocal of both sides to get K

K = k1k2k3 / (k2k3+k1k3+k1k2)