Two identical springs of spring constant 2k are attached to a block of mass m and to fixed support (

Q: Why are the springs considered to be in parallel? They look like they are in series because they are in a straight line!

This is a classic visual trap! In physics, springs are in parallel if they undergo the *same displacement* and their restoring forces add up. When you move the block by $x$, both springs deform by exactly $x$, and both push/pull the block in the same direction. Hence, they are mathematically in parallel.

Q: What would happen if the block was displaced to the left instead of the right?

The physics remains perfectly symmetric. The left spring would compress and push right, while the right spring would stretch and pull right. The net restoring force would still be $4kx$ directed towards the equilibrium position.

Q: Why did we add the spring constants instead of using the reciprocal formula?

The reciprocal formula $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2}$ is for springs in *series*, where they experience the same force but different displacements. Here, they experience the same displacement but exert different (or in this case, additive) forces, which is the definition of a parallel combination.

Q: Does the mass of the springs affect the time period?

In this standard JEE problem, we assume the springs are ideal and massless. If the springs had significant mass, we would have to add a fraction of their mass (usually $\frac{m_{spring}}{3}$) to the block's mass, which would increase the time period.

Q: Can I solve this using the energy method instead of the force method?

Absolutely! The energy method is incredibly elegant. The total potential energy when displaced by $x$ is $U = \frac{1}{2}(2k)x^2 + \frac{1}{2}(2k)x^2 = \frac{1}{2}(4k)x^2$. Comparing this to the standard $U = \frac{1}{2}k_{eq}x^2$, you immediately see that $k_{eq} = 4k$.

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. The time period of oscillations of this system is

Options:

(A) $2\pi\sqrt{\frac{m}{2k}}$

(B) $2\pi\sqrt{\frac{m}{k}}$

(C) $\pi\sqrt{\frac{m}{k}}$

(D) $\pi\sqrt{\frac{m}{2k}}$

Sigma Percentile · Accepted Answer

Imagine you are standing in a physics laboratory, looking at a fascinating mechanical setup. A block of mass $m$ is resting peacefully on a frictionless horizontal surface, perfectly balanced and sandwiched between two identical springs. Each of these springs has a spring constant of $2k$, and their other ends are firmly attached to rigid walls.

This is a classic simple harmonic motion (SHM) system, but it holds a beautiful conceptual trap that catches many students off guard. At first glance, because the springs are arranged in a straight line, your brain might scream, "They are in series!" But in physics, looks can be deceiving. We must always trust the forces. Let's dive into the mechanics of this system and uncover the truth.

### Analyzing the Setup

To truly understand any mechanical system, we must disturb its peace. Let's imagine grabbing the block and displacing it slightly to the right by a small distance $x$.

What happens to the springs? The spring on the left is forced to stretch by a distance $x$. According to Hooke's Law, it doesn't like being stretched, so it exerts a restoring force pulling the block back to the left.

Simultaneously, the spring on the right is forced to compress by that exact same distance $x$. It doesn't like being compressed either, so it exerts a restoring force pushing the block away, which is also to the left!

### The Parallel Illusion

This is the crucial "Aha!" moment. Both the stretched left spring and the compressed right spring are working together. They are both exerting a force on the block in the exact same direction—towards the equilibrium position.

In physics, when two springs undergo the same displacement and their restoring forces add up to pull or push the mass in the same direction, they are defined as being in a **parallel combination**. It doesn't matter that they are physically located on opposite sides of the block; dynamically, they are acting in parallel.

For springs in parallel, the equivalent spring constant $k_{eq}$ is simply the algebraic sum of their individual spring constants:
$$k_{eq} = k_1 + k_2$$

Since both springs in our system have a spring constant of $2k$, we can easily calculate the total stiffness of the system:
$$k_{eq} = 2k + 2k = 4k$$

The entire system behaves exactly like a single, much stiffer spring with a constant of $4k$ attached to our mass $m$.

### The Master Equation

Now that we have reduced our complex two-spring system into a simple one-spring system, we can bring in our master equation for the time period of a simple harmonic oscillator.

The time period $T$ of a spring-mass system is given by the elegant formula:
$$T = 2\pi\sqrt{\frac{m}{k_{eq}}}$$

This equation tells us that the time period depends directly on the inertia of the system (the mass $m$) and inversely on the restoring strength of the system (the equivalent spring constant $k_{eq}$).

### Final Calculation

We are now at the final step of our journey. We just need to substitute our calculated equivalent spring constant into the master equation.

Plugging in $k_{eq} = 4k$, we get:
$$T = 2\pi\sqrt{\frac{m}{4k}}$$

We can simplify this expression by taking the square root of the $4$ in the denominator, which gives us a $2$ outside the square root:
$$T = 2\pi \cdot \frac{1}{2} \sqrt{\frac{m}{k}}$$

The $2$ in the numerator and the $2$ in the denominator perfectly cancel each other out, leaving us with our final, beautifully simple answer:
$$T = \pi\sqrt{\frac{m}{k}}$$

And there we have it! By carefully analyzing the forces rather than just looking at the geometry, we avoided the series trap and arrived at the correct time period. Always remember: in physics, the direction of the restoring force is your ultimate guide.

Select Answer:

Visualized Solution

Visual Anchor

Displacement Analysis

Restoring Forces

Parallel Combination

Equivalent Spring Constant

Time Period Formula

Substitution

Final Answer

The Way Forward

The Sigma Insight: Simple Harmonic Motion (SHM)

Analyzing the Setup

The Parallel Illusion

The Master Equation

Final Calculation

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