Question

If two similar springs each of spring constant $K_1$ are joined in series, the new spring constant and time period would be changed by a factor:

• A. $\frac{1}{2},\sqrt{2}$
• B. $\frac{1}{4},2\sqrt{2}$
• C. $\frac{1}{2},2\sqrt{2}$
• D. $\frac{1}{4},\sqrt{2}$

Answer 1

The correct option is A

$\frac{1}{2},\sqrt{2}$

Explanation for the correct answer:

Step 1: Given

Step 1: Find the value of the new spring constant:

$\begin{array}{rcl}T& =& 2\pi \sqrt{\left(\frac{m}{K_{eq}}\right)}\end{array}$where, $m$ is the mass which is attached to these springs , $T$ is the time period and $K_{eq}$ is spring constant at equilibrium position.

Since, the springs are in series, then

$\begin{array}{rcl}\frac{1}{K_{eq}}& =& \frac{1}{K_1}+\frac{1}{K_2}\\ & \Rightarrow & \frac{1}{K_{eq}}=\frac{K_1+K_2}{K_1{K}_2}\\ & \Rightarrow & K_{eq}=\frac{K_1{K}_1}{2K_1}\left[\because K_1=K_2\right]\\ & \Rightarrow & K_{eq}=\frac{K_1}{2}\end{array}$

Step 2: Find the value of the new time period:

$\begin{array}{rcl}T\text{'}& =& 2\pi \sqrt{\left(\frac{m}{K_{eq}}\right)}\\ & =& 2\pi \sqrt{\left(\frac{m}{\left({\displaystyle \frac{K_1}{2}}\right)}\right)}\\ & =& 2\pi \sqrt{\left(\frac{2m}{\left({\displaystyle K_1}\right)}\right)}\\ \mathbf{[}\mathbf{\because }\mathbf{}\mathbf{T}& \mathbf{=}& \mathbf{2}\mathbf{\pi }\sqrt{\frac{\mathbf{m}}{{\mathbf{K}}_{\mathbf{1}}}}\mathbf{]}\\ & & \\ & =& \sqrt{2}\end{array}$

Hence, the correct option is A.