In the given figure, a body of mass M is held between two massless springs, on a smooth inclined pla

Q: Why are the springs considered to be in parallel and not in series? They look like they are in a single line!

Visual arrangement can be deceiving! In physics, springs are in parallel if they undergo the *same displacement* and their restoring forces *add up*. Here, displacing the block by $x$ changes the length of both springs by $x$, and both push/pull the block back to equilibrium. Hence, they are in parallel.

Q: Doesn't the angle of the incline $\alpha$ affect the time period?

No, it doesn't! A constant external force (like the component of gravity $Mg\sin\alpha$ down the incline) only shifts the mean equilibrium position where the net force is zero. It does not alter the restoring force gradient ($k_{eq}$), so the time period remains completely unaffected.

Q: What if the two springs had different spring constants, say $k_1$ and $k_2$?

The logic remains exactly the same. The equivalent spring constant would simply be $k_{eq} = k_1 + k_2$, and the frequency would be $f = \frac{1}{2\pi}\sqrt{\frac{k_1 + k_2}{M}}$.

Q: Why did we ignore the normal force and friction?

The problem explicitly states that the inclined plane is "smooth", which means friction is zero. The normal force acts perpendicular to the motion and balances the perpendicular component of gravity ($Mg\cos\alpha$), so it doesn't contribute to the restoring force along the incline.

Q: Would the frequency change if this entire setup was placed in a vertically accelerating elevator?

Surprisingly, no! An accelerating elevator would just change the effective gravity ($g_{eff}$). As we established earlier, constant forces (even if $g$ changes) only shift the equilibrium position but do not change the frequency of a spring-mass system.

Question

In the given figure, a body of mass $M$ is held between two massless springs, on a smooth inclined plane. The free ends of the springs are attached to firm supports. If each spring has spring constant $k$, then the frequency of oscillation of given body is

Options:

(A) $\frac{1}{2\pi}\sqrt{\frac{k}{M}}$

(B) $\frac{1}{2\pi}\sqrt{\frac{2k}{Mg\sin\alpha}}$

(C) $\frac{1}{2\pi}\sqrt{\frac{2k}{M}}$

(D) $\frac{1}{2\pi}\sqrt{\frac{k}{Mg\sin\alpha}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing next to a smooth inclined plane. Resting on this slope is a block of mass $M$.

This isn't just sliding down, though. It is held perfectly in equilibrium by two identical massless springs, each with a spring constant $k$. One spring anchors it to the top of the incline, and the other anchors it to the bottom.

At first glance, you might wonder how the angle of the incline, $\alpha$, or the pull of gravity affects the oscillation. Let's break down the physics step by step and uncover the elegant truth behind this system.

### The Magic of Parallel Springs

To understand how this system oscillates, we need to disturb it. Let's pull the block slightly down the incline by a small distance $x$ and freeze time.

What happens to the springs?

The bottom spring gets compressed by exactly $x$. According to Hooke's Law, it wants to expand back, so it pushes the block **up** the incline with a restoring force of $kx$.

Simultaneously, the top spring gets stretched by the exact same distance $x$. It wants to contract, so it pulls the block **up** the incline, also with a restoring force of $kx$.

**Notice the beautiful synergy here:** Both springs are working together! They are both forcing the block back towards its original equilibrium position.

Because their restoring forces add up in the same direction, these springs are effectively connected in **parallel**.

### The Master Equation

When springs are in parallel, their equivalent spring constant $k_{eq}$ is simply the sum of their individual spring constants.

$$k_{eq} = k + k = 2k$$

This means our two-spring system behaves exactly like a single, stiffer spring with a constant of $2k$.

Now, we bring in the master equation for the time period $T$ of a simple spring-mass system:

$$T = 2\pi \sqrt{\frac{M}{k_{eq}}}$$

### Final Calculation

We substitute our equivalent spring constant into the time period formula:

$$T = 2\pi \sqrt{\frac{M}{2k}}$$

But the question doesn't ask for the time period; it asks for the **frequency** of oscillation, $f$. Frequency is simply the reciprocal of the time period ($f = \frac{1}{T}$).

Flipping our equation, we get the final answer:

$$f = \frac{1}{2\pi} \sqrt{\frac{2k}{M}}$$

### The Gravity Illusion

You might be thinking, "Wait, what about gravity? What about the angle $\alpha$?"

This is a classic trap in physics! A constant external force—like the component of gravity acting down the incline ($Mg \sin\alpha$)—does absolutely nothing to the frequency of a spring-mass system.

All gravity does is stretch the top spring and compress the bottom spring initially to establish a new mean position where the net force is zero. Once that equilibrium is set, the restoring force for any *further* displacement $x$ depends purely on the springs ($2kx$).

The stiffness of the springs dictates the oscillation, making the frequency beautifully independent of gravity!

Select Answer:

Visualized Solution

$Initial Setup$

$Displacement Analysis$

$Restoring Force$

$Equivalent Spring Constant$

$Time Period$

$Frequency$

$Effect of Gravity$

The Sigma Insight: Simple Harmonic Motion (SHM)

Analyzing the Setup

The Magic of Parallel Springs

The Master Equation

Final Calculation

The Gravity Illusion

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Displacement Analysis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Restoring Force

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equivalent Spring Constant

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Time Period

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Frequency

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Effect of Gravity