A wooden block performs SHM on a frictionless surface with frequency v0. The block carries a charge

Q: Why doesn't the constant electric force change the frequency of oscillation?

The frequency of SHM depends on the *restoring force gradient* (how rapidly the force changes with displacement, i.e., $\frac{dF}{dx} = -k$). A constant force like $QE$ has a derivative of zero with respect to position, so it does not change the force gradient or the effective spring constant. Hence, the frequency remains exactly the same.

Q: What would happen if the electric field was alternating instead of uniform and constant?

If the electric field was alternating (e.g., $E = E_0 \sin(\omega_d t)$), it would act as a periodic driving force. This would result in *forced oscillations* rather than simple harmonic motion, and the block would eventually oscillate with the frequency of the alternating field, not its natural frequency.

Q: Does the amplitude of oscillation change when the electric field is switched on?

It depends on *when* and *how* the field is switched on. If the block is initially at rest at the unstretched position and the field is suddenly switched on, the block will find itself at a distance $x_0 = \frac{QE}{k}$ away from the new mean position. It will therefore start oscillating with an amplitude equal to $x_0$.

Q: Does this mean gravity also doesn't change the frequency of a vertical spring-mass system?

Exactly! When a spring-mass system is hung vertically, the constant gravitational force $mg$ only stretches the spring to a new equilibrium position by $x_0 = \frac{mg}{k}$. The frequency of vertical oscillation is still $\nu = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$, which is identical to its horizontal oscillation frequency.

Q: What if the charge on the block was negative instead of positive?

If the charge was negative ($-Q$), the constant electrostatic force would act to the left ($F_e = -QE$). The mean position would shift to the left by $x_0 = -\frac{QE}{k}$, but the frequency of oscillation would still remain completely unchanged at $\nu_0$.

Question

A wooden block performs SHM on a frictionless surface with frequency $
u_0$. The block carries a charge $+Q$ on its surface. If now a uniform electric field $\mathbf{E}$ is switched-on as shown, then the SHM of the block will be

Options:

(A) of the same frequency and with shifted mean position

(B) of the same frequency and with the same mean position

(C) of changed frequency and with shifted mean position

(D) of changed frequency and with the same mean position

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a classic spring-mass system resting on a perfectly smooth, frictionless horizontal table.

The block has a mass $m$, and it is attached to a rigid wall via a spring of force constant $k$.

Initially, the block is set into motion, and it happily oscillates back and forth about its natural, unstretched position of the spring.

This natural position serves as the **mean position** (or equilibrium position) of the oscillation, which we can define as $x = 0$.

The frequency of this simple harmonic motion (SHM) is a fundamental property of the system, given by:

$$
u_0 = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$$

This frequency is determined solely by the intrinsic parameters of the system: the stiffness of the spring ($k$) and the inertia of the block ($m$).

---

### Introducing the Electric Field

Now, let us introduce a twist to this setup.

We place a positive charge $+Q$ on the surface of the block and switch on a uniform horizontal electric field $\mathbf{E}$ pointing to the right.

Because the block is charged, it immediately experiences an electrostatic force:

$$F_e = QE$$

Since the electric field is uniform, this force is **constant in both magnitude and direction**, pulling the block steadily to the right.

How does this constant force affect the motion of the block?

---

### Finding the New Equilibrium (Mean) Position

To understand the new motion, we must locate the position where the net force on the block is zero.

This is the new equilibrium position, or the **mean position** of the oscillation.

Let $x$ be the displacement of the block from the original unstretched position of the spring.

The net force acting on the block at any position $x$ is the vector sum of the spring's restoring force and the constant electric force:

$$F_{	ext{net}} = -kx + QE$$

At the new equilibrium position $x = x_0$, the net force must be zero:

$$-kx_0 + QE = 0 \implies x_0 = \frac{QE}{k}$$

This tells us that the mean position is no longer at the unstretched length of the spring.

Instead, it has **shifted to the right** by a distance of $\frac{QE}{k}$.

---

### Analyzing the Dynamics about the New Mean Position

Now, let us analyze the motion of the block when it is displaced from this new equilibrium position.

Let $y$ represent the displacement of the block relative to the new mean position $x_0$.

Therefore, the total displacement from the unstretched position is:

$$x = x_0 + y$$

Let's substitute this expression back into our net force equation:

$$F_{	ext{net}} = -k(x_0 + y) + QE$$

$$F_{	ext{net}} = -kx_0 - ky + QE$$

Recall that from our equilibrium condition, $kx_0 = QE$.

Substituting this in, we see a beautiful cancellation:

$$F_{	ext{net}} = -QE - ky + QE = -ky$$

Using Newton's second law, we can write the equation of motion as:

$$m \frac{d^2y}{dt^2} = -ky$$

This is the exact, standard differential equation for simple harmonic motion!

---

### Conclusion and Key Takeaway

The resulting motion is still simple harmonic, and its angular frequency is:

$$\omega = \sqrt{\frac{k}{m}}$$

The frequency of oscillation is:

$$
u = \frac{1}{2\pi} \sqrt{\frac{k}{m}} = 
u_0$$

Thus, the frequency of oscillation remains **completely unchanged**, while the mean position is **shifted to the right** by $\frac{QE}{k}$.

This leads us directly to **Option (a)**.

**The Golden Rule of SHM:** Any constant external force acting on a simple harmonic oscillator (such as gravity, a constant electric force, or a steady wind) **only shifts the mean position** of the oscillation. It **never** alters the frequency or time period, because it does not change the force gradient (the stiffness $k$) of the system.

Animated Solution for Physics - Oscillations: A wooden block performs SHM on a frictionless surface with frequency $ν_{0}$ . The block carries a charge $+ Q$ on its surface. If now a uniform electric field $E$ is switched-on as shown, then the SHM of the block will be

Select Answer:

Visualized Solution

Visualizing the Setup

The Natural Frequency of SHM

Switching on the Electric Field

The New Equation of Motion

Locating the New Mean Position

Calculating the Shift $x_{0}$

Equation of Motion about New Mean

Simplifying the Force Equation

The New Frequency of Oscillation

The Way Forward: Constant Forces in SHM

The Sigma Insight: Force and Energy Method in SHM

Analyzing the Setup

Introducing the Electric Field

Finding the New Equilibrium (Mean) Position

Analyzing the Dynamics about the New Mean Position

Conclusion and Key Takeaway

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