Question

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.

Answer 1

Answer: Equivalent spring constant
`{k_(eq)}=(1)/(k_(1))+(1)/(k_(2))`

Explanation: Spring constant, also known as stiffness constant is the amount of load (force) required to make a deformation of unit length in the size of the spring from its free .

Mathematically given as:

`F=-kx` ...............(1)

In series the springs are connected back to back consecutively in a single line as shown in figure.

In such a case:

• The equivalent elongation
  `x_(eq)` is the sum of the respective elongations of each spring.
• The equivalent force
  `F_(eq)` is equal in each spring.

Mathematically represented as:

`x_(eq)=x_(1)+x_(2)` ...........................(2)

∵ From eq. (1) & (2) we have,

`x_(eq)=x_(1)+x_(2)` ∵F=-kx, where negative sign denotes that the restoration force acts in the direction opposite to the applied force.

`\Rightarrow (F_(eq))/(k_(eq))=(F_(1))/(k_(1))+(F_(2))/(k_(2))`

∵The equivalent force is equal on each spring.

We get the relation as:

`\Rightarrow(1)/(k_(eq))=(1)/(k_(1))+(1)/(k_(2))`