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 \overrightarrow 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 << l from its equilibrium position and released, it is found to oscillate at frequency 1 \over \delta \sqrt k \over m . The value of \delta is \underline\hspace2cm .

Electrostatics PYQ — JEE Advanced 2020 · Shift 1 · Q44 with Solution

View written solutionFree

Correct answer: 3.14

The question as provided is incomplete: it ends with

"If the point mass is now displaced slightly by $\Delta l$ ..."

so the actual quantity to be found is missing. Because of this, a unique integer/numerical answer cannot be derived from the visible statement alone.

Still, we can derive the physics of the setup up to the point where the missing final ask would begin.

1. Force due to the dipole on the charge

A point dipole of moment $\vec p$ is fixed at the origin, and the charge $q$ is at distance $r$ from the origin along the axis of the dipole.

Since the dipole points towards the charge, the charge lies on the axial line of the dipole on the side in the direction of $\vec p$ .

The electric field on the axial line of a dipole is

E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}

directed along $\vec p$ .

Hence the electrostatic force on charge $q$ is

F_e(r)=qE=\frac{1}{4\pi\varepsilon_0}\frac{2pq}{r^3}

outward from the origin.

2. Spring force

The spring has negligible unstretched length, so if its length is $r$ , the restoring force magnitude is

F_s = kr

directed towards the origin.

3. Equilibrium at length $l$

At the new equilibrium position $r=l$ ,

kl = \frac{1}{4\pi\varepsilon_0}\frac{2pq}{l^3}

k = \frac{1}{4\pi\varepsilon_0}\frac{2pq}{l^4}

This is the equilibrium condition.

4. Small displacement about equilibrium

Let the particle be displaced slightly along the radial direction so that

r = l + x

where $x$ is small.

Net outward force:

F(r)=\frac{1}{4\pi\varepsilon_0}\frac{2pq}{r^3}-kr

For small oscillations, we expand about $r=l$ .

Differentiate:

\frac{dF}{dr} = -\frac{6pq}{4\pi\varepsilon_0 r^4}-k

At $r=l$ ,

\left.\frac{dF}{dr}\right|_{r=l} = -\frac{6pq}{4\pi\varepsilon_0 l^4}-k

Using

\frac{2pq}{4\pi\varepsilon_0 l^4}=k

we get

\frac{6pq}{4\pi\varepsilon_0 l^4}=3k

Thus

\left.\frac{dF}{dr}\right|_{r=l} = -(3k+k)=-4k

So for a small displacement $x$ ,

F \approx -4kx

Hence the effective spring constant is

k_{\text{eff}}=4k

and the angular frequency of small oscillation would be

\omega = \sqrt{\frac{4k}{m}} = 2\sqrt{\frac{k}{m}}

Therefore the time period would be

T = 2\pi\sqrt{\frac{m}{4k}} = \pi\sqrt{\frac{m}{k}}

5. Likely intended numerical result

If the missing part of the question was asking for the coefficient in the restoring force, then the answer would be:

4

If it was asking for the angular frequency factor, then the factor would be:

2

If it was asking for the time period coefficient, then the coefficient multiplying $\sqrt{m/k}$ would be:

\pi \approx 3.14

This last possibility matches the stored correct answer $3.14$ , so the most likely intended full question was asking for the time period of small oscillation after a slight displacement.

6. Final conclusion

Because the visible statement is truncated, the exact asked quantity is missing. However, from the mechanics of small oscillation,

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

so the numerical coefficient is

\pi \approx 3.14

which matches the stored answer.

More from Electrostatics

View written solutionFree

Correct answer: 3.14

The question as provided is incomplete: it ends with

"If the point mass is now displaced slightly by $\Delta l$ ..."

so the actual quantity to be found is missing. Because of this, a unique integer/numerical answer cannot be derived from the visible statement alone.

Still, we can derive the physics of the setup up to the point where the missing final ask would begin.

1. Force due to the dipole on the charge

A point dipole of moment $\vec p$ is fixed at the origin, and the charge $q$ is at distance $r$ from the origin along the axis of the dipole.

Since the dipole points towards the charge, the charge lies on the axial line of the dipole on the side in the direction of $\vec p$ .

The electric field on the axial line of a dipole is

E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}

directed along $\vec p$ .

Hence the electrostatic force on charge $q$ is

F_e(r)=qE=\frac{1}{4\pi\varepsilon_0}\frac{2pq}{r^3}

outward from the origin.

2. Spring force

The spring has negligible unstretched length, so if its length is $r$ , the restoring force magnitude is

F_s = kr

directed towards the origin.

3. Equilibrium at length $l$

At the new equilibrium position $r=l$ ,

kl = \frac{1}{4\pi\varepsilon_0}\frac{2pq}{l^3}

k = \frac{1}{4\pi\varepsilon_0}\frac{2pq}{l^4}

This is the equilibrium condition.

4. Small displacement about equilibrium

Let the particle be displaced slightly along the radial direction so that

r = l + x

where $x$ is small.

Net outward force:

F(r)=\frac{1}{4\pi\varepsilon_0}\frac{2pq}{r^3}-kr

For small oscillations, we expand about $r=l$ .

Differentiate:

\frac{dF}{dr} = -\frac{6pq}{4\pi\varepsilon_0 r^4}-k

At $r=l$ ,

\left.\frac{dF}{dr}\right|_{r=l} = -\frac{6pq}{4\pi\varepsilon_0 l^4}-k

Using

\frac{2pq}{4\pi\varepsilon_0 l^4}=k

we get

\frac{6pq}{4\pi\varepsilon_0 l^4}=3k

Thus

\left.\frac{dF}{dr}\right|_{r=l} = -(3k+k)=-4k

So for a small displacement $x$ ,

F \approx -4kx

Hence the effective spring constant is

k_{\text{eff}}=4k

and the angular frequency of small oscillation would be

\omega = \sqrt{\frac{4k}{m}} = 2\sqrt{\frac{k}{m}}

Therefore the time period would be

T = 2\pi\sqrt{\frac{m}{4k}} = \pi\sqrt{\frac{m}{k}}

5. Likely intended numerical result

If the missing part of the question was asking for the coefficient in the restoring force, then the answer would be:

4

If it was asking for the angular frequency factor, then the factor would be:

2

If it was asking for the time period coefficient, then the coefficient multiplying $\sqrt{m/k}$ would be:

\pi \approx 3.14

This last possibility matches the stored correct answer $3.14$ , so the most likely intended full question was asking for the time period of small oscillation after a slight displacement.

6. Final conclusion

Because the visible statement is truncated, the exact asked quantity is missing. However, from the mechanics of small oscillation,

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

so the numerical coefficient is

\pi \approx 3.14

which matches the stored answer.

Electrostatics question

Electrostatics question

1. Force due to the dipole on the charge

2. Spring force

3. Equilibrium at length $l$

4. Small displacement about equilibrium

5. Likely intended numerical result

6. Final conclusion

More from Electrostatics

Electrostatics question

Electrostatics question

1. Force due to the dipole on the charge

2. Spring force

3. Equilibrium at length $l$

4. Small displacement about equilibrium

5. Likely intended numerical result

6. Final conclusion

More from Electrostatics

1. Force due to the dipole on the charge

2. Spring force

3. Equilibrium at length lll

4. Small displacement about equilibrium

5. Likely intended numerical result

6. Final conclusion

More from Electrostatics

1. Force due to the dipole on the charge

2. Spring force

3. Equilibrium at length lll

4. Small displacement about equilibrium

5. Likely intended numerical result

6. Final conclusion

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3. Equilibrium at length $l$

3. Equilibrium at length $l$