One end of a spring of negligible unstretched length and spring constant k is fixed at the origin (0

Q: Why do we assume the unstretched length is zero?

The problem explicitly states 'negligible unstretched length', which means the extension of the spring is simply its total length $\ell$. This simplifies the spring force to $k\ell$.

Q: Why do we use the binomial approximation here?

Because the displacement $x$ (or $\Delta\ell$) is given as much smaller than the equilibrium length $\ell$. This allows us to linearize the complex electrostatic force, which is a standard technique for finding SHM frequencies.

Q: Why did the $k\ell$ term disappear in the final force equation?

At the equilibrium position, the net force is zero, meaning $k\ell = F_{\text{electric}}$. When we expand the perturbed force, the constant terms represent this equilibrium state and naturally cancel out, leaving only the terms proportional to $x$.

Q: Why is the effective spring constant $4k$ and not just $k$?

When you displace the charge, the spring pulls it back with an extra force $kx$, but the electrostatic repulsion also weakens, effectively 'helping' the spring push the charge back. This gradient of the electric field adds an extra $3k$ to the stiffness.

Q: Why did JEE accept two different answers for this question?

The term 'frequency' can refer to either angular frequency ($\omega$) or linear frequency ($f$). Depending on which one you assume, $\delta$ evaluates to either $0.5$ or $\pi$ ($3.14$). Both interpretations are technically valid.

Question

One end of a spring of negligible unstretched length and spring constant $k$ is fixed at the origin $(0,0)$. A point particle of mass $m$ carrying a positive charge $q$ is attached at its other end. The entire system is kept on a smooth horizontal surface. When a point dipole $\vec{p}$ pointing towards the charge $q$ is fixed at the origin, the spring gets stretched to a length $\ell$ and attains a new equilibrium position (see figure below). If the point mass is now displaced slightly by $\Delta\ell \ll \ell$ from its equilibrium position and released, it is found to oscillate at frequency $\frac{1}{\delta}\sqrt{\frac{k}{m}}$. The value of $\delta$ is _________.

Sigma Percentile · Accepted Answer

### The Setup: A Delicate Balance

Imagine a beautifully balanced physical system. We have a spring with a negligible unstretched length, meaning its restoring force is simply proportional to its total length, $F_{	ext{spring}} = k\ell$. At the end of this spring sits a point mass $m$ carrying a charge $q$.

At the origin, a dipole $\vec{p}$ is fixed, pointing directly at our charge. This dipole creates an electric field along its axis, exerting an outward electrostatic repulsive force on the charge. In the equilibrium state, these two opposing forces are locked in a perfect stalemate.

Mathematically, we can write this balance as:

$$k\ell = \frac{1}{4\pi\epsilon_0} \frac{2pq}{\ell^3}$$

Let's define $K = \frac{1}{4\pi\epsilon_0}$ to keep our equations clean. So, $k\ell = \frac{2Kpq}{\ell^3}$. This equation is our anchor; it defines the geometry of the equilibrium.

### The Perturbation: Disturbing the Peace

Now, let's disturb this peace. We displace the mass by a tiny distance $x$ (which the problem calls $\Delta\ell$). The forces are no longer balanced, and a net restoring force will act on the particle.

As the spring stretches further to a length of $(\ell + x)$, its inward pull increases. Simultaneously, the charge moves further away from the dipole, so the outward electrostatic repulsion weakens. Both of these changes work together to pull the mass back towards equilibrium.

The new net restoring force is:

$$F_{	ext{net}} = k(\ell + x) - \frac{2Kpq}{(\ell + x)^3}$$

### The Math Magic: Binomial Approximation

Here is the catch. The displacement $x$ is extremely small compared to $\ell$ ($x \ll \ell$). This is a classic setup for the binomial approximation. We factor out $\ell^3$ from the denominator of the electrostatic term and move it to the numerator:

$$F_{	ext{net}} = k(\ell + x) - \frac{2Kpq}{\ell^3(1 + x/\ell)^3}$$

Using the approximation $(1 + y)^{-n} \approx 1 - ny$ for very small $y$, we get:

$$F_{	ext{net}} \approx k(\ell + x) - \frac{2Kpq}{\ell^3}\left(1 - \frac{3x}{\ell}ight)$$

Let's open the brackets carefully. Notice how the initial equilibrium terms perfectly cancel each other out. The $k\ell$ from the spring force cancels exactly with the $-\frac{2Kpq}{\ell^3}$ from the electrostatic force.

We are left with:

$$F_{	ext{net}} = kx + \frac{2Kpq}{\ell^3}\left(\frac{3x}{\ell}ight)$$

Since we know from our equilibrium condition that $\frac{2Kpq}{\ell^3} = k\ell$, we can substitute this back in:

$$F_{	ext{net}} = kx + k\ell\left(\frac{3x}{\ell}ight) = kx + 3kx = 4kx$$

### The Ambiguity: Angular vs. Linear Frequency

Our effective spring constant $k_{	ext{eq}}$ is $4k$. The gradient of the electric field effectively added a stiffness of $3k$ to our system!

Now, there is a slight ambiguity in the question's wording. It asks for the 'frequency'.

If it means **angular frequency** ($\omega$):

$$\omega = \sqrt{\frac{k_{	ext{eq}}}{m}} = \sqrt{\frac{4k}{m}} = 2\sqrt{\frac{k}{m}}$$

Comparing this to $\frac{1}{\delta}\sqrt{\frac{k}{m}}$, we get $\delta = 0.50$.

If it means **linear frequency** ($f$):

$$f = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{4k}{m}} = \frac{1}{\pi}\sqrt{\frac{k}{m}}$$

Comparing this, we get $\delta = \pi \approx 3.14$.

Because of this dual interpretation, JEE Advanced accepted both $0.50$ and the range $3.13 - 3.15$ as correct answers. A beautiful problem that tests both your physics intuition and your mathematical rigor!

Enter Numerical Value:

Visualized Solution

$F_{net} = 0$

$Δ x \neq = 0$

$F_{net}$

$(1 + x)^{n} \approx 1 + n x$

$k_{eq}$

$δ$

\text{Extensions}

The Sigma Insight: Force and Energy Method in SHM

The Setup

The Perturbation

The Math Magic

The Ambiguity

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Enter Numerical Value:

Visualized Solution

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