A block of mass 2M is attached to a massless spring with spring-constant k. This block is connected

Q: Why is the tension pulling the movable pulley $2T$?

The movable pulley is considered massless. For a massless object to have a finite acceleration, the net force acting on it must be zero. The string going over it pulls down with tension $T$ on both sides, creating a total downward force of $2T$. Therefore, the string pulling it upwards must exert a tension of exactly $2T$ to balance it.

Q: How did we get the kinematic constraint $a_2 + a_3 = 2a_1$?

Imagine the movable pulley moves down by a distance $x_1$. The total length of the string draped over it is $l = (x_2 - x_1) + (x_3 - x_1)$, where $x_2$ and $x_3$ are the absolute downward displacements of the hanging blocks. Since the string length $l$ is constant, differentiating this equation twice with respect to time gives $0 = a_2 - a_1 + a_3 - a_1$, which simplifies to $a_2 + a_3 = 2a_1$.

Q: Why is the initial position considered an extreme position?

The problem states the system is released from rest. In any oscillatory motion, the points where the velocity is momentarily zero are the extreme positions (the turning points). Since it starts from rest at the unstretched length ($x=0$), this point must be one of the extremes of the oscillation.

Q: Why is the maximum extension exactly $2x_{eq}$?

In Simple Harmonic Motion, the motion is perfectly symmetric around the equilibrium position ($x_{eq}$). If the oscillation starts at $x=0$, the amplitude is the distance from the start to the equilibrium, which is $x_{eq} - 0 = x_{eq}$. The block will swing past the equilibrium by this same amplitude to reach the other extreme, making the maximum extension $x_{eq} + A = 2x_{eq}$.

Q: Can we solve this problem using energy conservation instead of forces?

You can use the work-energy theorem to find the maximum extension and the maximum velocity. However, to find the acceleration at a specific point or to rigorously prove that the motion is Simple Harmonic, you must use the force method to derive the equation of motion.

Question

A block of mass 2M is attached to a massless spring with spring-constant k. This block is connected to two other blocks of masses M and 2M using two massless pulleys and strings. The accelerations of the blocks are $a_1$, $a_2$ and $a_3$ as shown in figure. The system is released from rest with the spring in its unstretched state. The maximum extension of the spring is $x_0$. Which of the following option(s) is/are correct?[g is the acceleration due to gravity. Neglect friction]

Options:

(A) $x_0 = \frac{4Mg}{k}$

(B) When spring achieves an extension of $\frac{x_0}{2}$ for the first time, the speed of the block connected to the spring is $3g\sqrt{\frac{M}{5k}}$

(C) $a_2 - a_1 = a_1 - a_3$

(D) At an extension of $\frac{x_0}{4}$ of the spring, the magnitude of acceleration of the block connected to the spring is $\frac{3g}{10}$

Sigma Percentile · Accepted Answer

### The Setup: A Symphony of Interconnected Masses

Imagine a beautifully orchestrated mechanical dance. On a frictionless table, a block of mass $2M$ is tethered to a wall by a spring of constant $k$. But it doesn't just oscillate freely; it's connected via a string to a movable pulley. This movable pulley, in turn, acts as the stage for two hanging blocks of masses $M$ and $2M$.

When the system is released from rest, the spring begins to stretch, the table block accelerates to the right, the movable pulley descends, and the hanging blocks undergo their own relative motion. To unravel this complex behavior, we must first understand the kinematic constraints binding them together.

### The Kinematic Dance: Tying the Accelerations Together

Let's focus on the movable pulley. If the block on the table moves to the right by a distance $x_1$, the movable pulley must move downwards by the exact same distance $x_1$. Now, consider the string draped over this movable pulley. The total length of this string remains constant.

If the left block ($M$) moves down by an absolute distance $x_2$ and the right block ($2M$) moves down by an absolute distance $x_3$, their displacements relative to the movable pulley are $(x_2 - x_1)$ and $(x_3 - x_1)$. Since the string length is constant, the sum of these relative displacements must be zero (assuming one moves up relative to the pulley and the other moves down, but mathematically, the average of their absolute downward displacements equals the pulley's displacement):

$$x_2 + x_3 = 2x_1$$

Differentiating this equation twice with respect to time yields the golden kinematic constraint for their accelerations:

$$a_2 + a_3 = 2a_1$$

By simply rearranging this equation, we get $a_2 - a_1 = a_1 - a_3$. This immediately confirms that **Option (C) is correct!** But we are physicists, and our journey doesn't end here. We must explore the dynamics.

### The Force Play: Newton's Laws and Tension

Let's analyze the forces. Let the tension in the string connecting the hanging blocks be $T$. Because the movable pulley is considered massless, it cannot sustain a net force. The downward pull from the two segments of the string is $T + T = 2T$. Therefore, the upward string—which connects to the table block—must exert a tension of $2T$.

Now, we write Newton's Second Law for each of the three blocks:

1.  **For the hanging mass $M$:** Gravity pulls down, tension pulls up.
    $$Mg - T = M a_2$$
2.  **For the hanging mass $2M$:**
    $$2Mg - T = 2M a_3$$
3.  **For the block on the table ($2M$):** It is pulled to the right by tension $2T$ and to the left by the spring force $kx$.
    $$2T - kx = 2M a_1$$

### The SHM Revelation: Unmasking the Oscillator

Our goal is to find the acceleration $a_1$ of the table block as a function of its position $x$. We can do this by eliminating $a_2$ and $a_3$. From the hanging blocks' equations, we isolate their accelerations:

$$a_2 = g - \frac{T}{M}$$
$$a_3 = g - \frac{T}{2M}$$

Substituting these into our kinematic constraint $2a_1 = a_2 + a_3$:

$$2a_1 = \left(g - \frac{T}{M}\right) + \left(g - \frac{T}{2M}\right) = 2g - \frac{3T}{2M}$$

Solving for the tension $T$ gives us a beautiful expression entirely in terms of $a_1$:

$$T = \frac{4M}{3}(g - a_1)$$

Now, we substitute this tension back into the equation for the table block:

$$2\left[\frac{4M}{3}(g - a_1)\right] - kx = 2M a_1$$

Expanding and grouping the $a_1$ terms together:

$$\frac{8Mg}{3} - kx = \left(2M + \frac{8M}{3}\right) a_1 = \frac{14M}{3} a_1$$

Isolating $a_1$, we arrive at the master equation of motion:

$$a_1 = -\frac{3k}{14M}\left(x - \frac{8Mg}{3k}\right)$$

Look closely at this structure. It is of the exact form $a = -\omega^2(x - x_{eq})$. This proves that the block on the table executes **Simple Harmonic Motion (SHM)**!

From this, we can extract the vital parameters of the oscillation:
*   **Equilibrium Position:** $x_{eq} = \frac{8Mg}{3k}$
*   **Angular Frequency Squared:** $\omega^2 = \frac{3k}{14M}$

### Reaping the Rewards: Evaluating the Options

Armed with the knowledge that the system performs SHM, we can easily evaluate the remaining options.

**Maximum Extension ($x_0$):**
The system is released from rest at the unstretched position ($x = 0$). In any oscillation, points of zero velocity are the extreme positions. Thus, the amplitude $A$ is the distance from the equilibrium position to this starting point:

$$A = x_{eq} - 0 = \frac{8Mg}{3k}$$

The maximum extension $x_0$ will be at the other extreme, which is twice the equilibrium distance:

$$x_0 = 2x_{eq} = \frac{16Mg}{3k}$$

This makes Option (A) incorrect.

**Speed at $x_0/2$:**
The position $x_0/2$ is exactly the equilibrium position $x_{eq}$. In SHM, the speed is maximum at the mean position, given by $v_{max} = A\omega$. Let's calculate it:

$$v_{max} = \left(\frac{8Mg}{3k}\right) \sqrt{\frac{3k}{14M}} = 8Mg \sqrt{\frac{1}{42kM}} = 8g\sqrt{\frac{M}{42k}}$$

This does not match the value in Option (B), making it incorrect.

**Acceleration at $x_0/4$:**
The position $x = x_0/4$ corresponds to $x_{eq}/2$. The displacement from the mean position is $X = x - x_{eq} = -A/2$. The acceleration at this point is:

$$a_1 = -\omega^2 X = -\left(\frac{3k}{14M}\right)\left(-\frac{4Mg}{3k}\right) = \frac{4g}{14} = \frac{2g}{7}$$

This is not $3g/10$, so Option (D) is also incorrect.

In conclusion, the intricate dance of these blocks is governed by a beautiful underlying harmonic rhythm, and only the kinematic constraint presented in **Option (C)** holds true.

Select Answer:

Visualized Solution

$S y s t e m O v er v i e w$

$K in e ma t i c C o n s t r ain t$

$T e n s i o n D i s t r ib u t i o n$

$N e w t o n^{'} s S eco n d L a w$

$S o l v in g f or T e n s i o n$

$E q u a t i o n o f M o t i o n f or 2 M$

$S H M C ha r a c t er i s t i cs$

$M a x im u m E x t e n s i o n (x_{0})$

$S p ee d a t M e an P os i t i o n$

$A cce l er a t i o n a t x_{0} /4$

The Sigma Insight: Force and Energy Method in SHM

The Setup

The Kinematic Dance

The Force Play

The SHM Revelation

Reaping the Rewards

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Visualized Solution

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Newton′s Second Law

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of Motion for 2M

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Maximum Extension (x0​)

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Acceleration at x0​/4