Question

Two identical springs of spring constant $‘2k’$ are attached to a block of mass $m$ and to fixed support (see figure). When the mass is displaced from equilibrium position on either side, it executes simple harmonic motion. Then, time period of oscillations of this system is:

• A. $\pi \sqrt{\frac{m}{k}}$
• B. $\pi \sqrt{\frac{m}{2k}}$
• C. $2\pi \sqrt{\frac{m}{k}}$
• D. $2\pi \sqrt{\frac{m}{2k}}$

Answer 1

The correct option is A

$\pi \sqrt{\frac{m}{k}}$

Step 1: Given

Spring constant of first spring: $k_1=2k$

Spring constant of second spring: $k_2=2k$

Mass of the block is $m$

Step 2: Formula Used

The time period of oscillation of a spring in SHM is given as

$T=2\pi \sqrt{\frac{m}{k}}$

Where, $m$ is mass and $k$ is spring constant.

Step 3: Find the time period of oscillation of system

Calculate the effective spring constant of the system by adding the spring constants, since the springs are connected in parallel combination

$\begin{array}{rcl}{k}_{eff}& =& k_1+k_2\\ & =& 2k+2k\\ & =& 4k\end{array}$

Calculate the time period of oscillation using the formula for time period.

$\begin{array}{rcl}T& =& 2\pi \sqrt{\frac{m}{k}}\\ & =& 2\pi \sqrt{\frac{m}{4k}}\\ & =& \frac{2\pi }{2}\sqrt{\frac{m}{k}}\\ & =& \pi \sqrt{\frac{m}{k}}\end{array}$

Hence, option A is correct.