A rigid dipole undergoes a simple harmonic motion about its centre in the presence of an electric field E1 = E0 x. If another electric field E2 = 2E0 (y + z) is introduced to the system, what will be the percentage change in the frequency of the oscillation (approximate)?

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q36 – JEE-MAIN 2026

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Home Physics Electric Charges and FieldsQuestion #97

A rigid dipole undergoes a simple harmonic motion about its centre in the presence of an electric field

{\vec{E}}_{1} = E_{0} \hat{x}

. If another electric field

{\vec{E}}_{2} = 2 E_{0} (\hat{y} + \hat{z})

is introduced to the system, what will be the percentage change in the frequency of the oscillation (approximate)?

Clue 1 of 3: Simple Harmonic Motion of a Dipole

Recall that a dipole in a uniform electric field experiences a restoring torque when displaced from its equilibrium orientation, leading to simple harmonic motion.

🧠 Full Solution Path

Step 1: Determine the initial frequency of oscillation✦ Active

The initial electric field is ${\vec{E}}_{1} = E_{0} \hat{x}$ , so its magnitude is $E_{1} = E_{0}$ . The frequency of oscillation for a dipole in an electric field is $f = \frac{1}{2 π} \sqrt{\frac{p E}{I}}$ , where $p$ is the dipole moment and $I$ is the moment of inertia. Thus, the initial frequency is:

f_{1} = \frac{1}{2 π} \sqrt{\frac{p E_{0}}{I}}

Step 2: Calculate the total electric field and the new frequency○ Expand

When the second electric field ${\vec{E}}_{2} = 2 E_{0} (\hat{y} + \hat{z})$ is introduced, the total electric field is the vector sum:

{\vec{E}}_{t o t a l} = {\vec{E}}_{1} + {\vec{E}}_{2} = E_{0} \hat{x} + 2 E_{0} (\hat{y} + \hat{z}) = E_{0} (\hat{x} + 2 \hat{y} + 2 \hat{z})

The magnitude of the total electric field is:

E_{t o t a l} = E_{0} \sqrt{1^{2} + 2^{2} + 2^{2}} = E_{0} \sqrt{1 + 4 + 4} = E_{0} \sqrt{9} = 3 E_{0}

The new frequency of oscillation $f_{2}$ will be:

f_{2} = \frac{1}{2 π} \sqrt{\frac{p E_{t o t a l}}{I}} = \frac{1}{2 π} \sqrt{\frac{p (3 E_{0})}{I}}

Step 3: Calculate the percentage change in frequency○ Expand

The ratio of the new frequency to the original frequency is:

\frac{f_{2}}{f_{1}} = \frac{\frac{1}{2 π} \sqrt{\frac{3 p E_{0}}{I}}}{\frac{1}{2 π} \sqrt{\frac{p E_{0}}{I}}} = \sqrt{3}

The percentage change in frequency is given by:

Percentage change = (\frac{f_{2}}{f_{1}} - 1) \times 100 % = (\sqrt{3} - 1) \times 100 %

Using the approximate value $\sqrt{3} \approx 1.732$ , the percentage change is:

(1.732 - 1) \times 100 % = 0.732 \times 100 % = 73.2 %

Approximately, the percentage change is $73 %$ . This corresponds to option 1.

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Physics Question 36 – JEE-MAIN 2026

Clue 1 of 3: Simple Harmonic Motion of a Dipole

🧠 Full Solution Path