JEE Advanced 2020 · Paper 1 · Physics Q15

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PhysicsNumerical+4Question 15 of 54

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 $l$ and attains a new equilibrium position (see figure below). If the point mass is now displaced slightly by $\Delta l \ll l$ from its equilibrium position and released, it is found to oscillate at frequency $\dfrac1\delta\sqrt{\dfrac km}$ . The value of $\delta$ is ______.

x-y axes. A dipole p at the origin points along +x toward a charge q; a spring stretched to length l joins the origin to q on the x-axis.

Your answer

Answer and worked solution

Official answer: 0.5

Forces on the charge at distance $x$ along the dipole's axis:

dipole field $E=\dfrac{2kp}{x^3}$ (pointing away from the dipole), so a force $+\dfrac{2kpq}{x^3}$ outward;
spring force $-kx$ (zero natural length).

Equilibrium at $x=l$ : $kl=\dfrac{2kpq}{l^3}$ , so $\dfrac{2kpq}{l^4}=k$ .

Stiffness about equilibrium.
$-\frac{dF}{dx}=k+\frac{6kpq}{l^4}=k+3k=4k.$
So $\omega=\sqrt{4k/m}=2\sqrt{k/m}$ and
$f=\frac\omega{2\pi}=\frac1\pi\sqrt{\frac km}\implies\delta=\pi\approx3.14.$

Answer: 3.14 (key: 3.13 to 3.15, or 0.50 if "frequency" is read as the angular frequency $\omega=2\sqrt{k/m}$ , which gives $\delta=0.5$ )

Topics:Spring-mass systems Electric dipole

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JEE Advanced 2020 · Paper 1, Physics question 15