A mass m is undergoing SHM in the vertical direction about the mean position y0 with amplitude A and

Q: Why does the block detach from the spring?

The block detaches when the spring starts to pull it downwards (or when the normal contact force/tension becomes zero). In this idealized problem, we assume detachment occurs at a specific displacement $y$ where we choose to release or uncouple it to maximize height.

Q: What if $A\omega^2 \le g$?

If $A\omega^2 \le g$, the maximum acceleration of the SHM is less than or equal to $g$. In this case, the block cannot reach the displacement $y = g/\omega^2$ because it is greater than or equal to the amplitude $A$. The maximum height would simply be achieved at the highest point of the SHM, i.e., $y = A$.

Q: Why is the acceleration after detachment equal to $-g$?

Once detached, the block is no longer connected to the spring, so the spring force becomes zero. The only force acting on the block is gravity, which acts downwards, causing a constant downward acceleration of $g$.

Q: Can we use conservation of energy to solve this?

Yes! The total energy at detachment is the sum of kinetic and potential energy relative to the mean position. After detachment, this energy is converted entirely into gravitational potential energy at the maximum height. Differentiating this energy relation yields the same result.

Q: Is the maximum height measured from the mean position or the detachment point?

The question asks for the distance $y$ measured from the mean position, but the height $h$ is also defined relative to the mean position. Our formulation $h = y + v^2/(2g)$ correctly measures the total height from the mean position.

Question

A mass $m$ is undergoing SHM in the vertical direction about the mean position $y_0$ with amplitude $A$ and angular frequency $\omega$. At a distance $y$ from the mean position, the mass detaches from the spring. Assume that the spring contracts and does not obstruct the motion of $m$. Find the distance $y$ (measured from the mean position) such that the height $h$ attained by the block is maximum. ($A\omega^2 > g$).

Sigma Percentile · Accepted Answer

### Introduction: The Intersection of SHM and Free Fall

Imagine a mass suspended from a vertical spring, dancing in a rhythmic, periodic motion. This is the classic **Simple Harmonic Motion (SHM)**.

But what happens if we suddenly cut the cord—or in this case, detach the mass from the spring mid-flight?

Suddenly, the comforting, restoring force of the spring vanishes. The mass is thrust into a new physical reality: **free fall under gravity**.

This problem asks us to find the optimal point of detachment $y$ from the mean position such that the total height $h$ reached by the block is maximized. It is a beautiful blend of SHM kinematics, projectile motion, and calculus-based optimization.

---

### Analyzing the Setup

Let's define our coordinate system. We measure all vertical displacements from the **mean position** $y_0$ of the SHM.

At any displacement $y$ from this mean position, the block has a certain velocity $v$.

If the block detaches at this point, it leaves the spring with this instantaneous velocity $v$ and begins to move upwards as a free particle under the sole influence of gravity.

Our goal is to maximize the total height $h$ attained above the mean position. This total height is composed of two parts:
1. The height $y$ at which the detachment occurs.
2. The additional height $h_{	ext{free}}$ reached during the free-fall phase.

$$
1.5h_1 = h_2
$$

---

### The Master Equation

First, let's write down the velocity of the block at displacement $y$ during SHM:

$$
v = \omega \sqrt{A^2 - y^2}
$$

Squaring both sides gives:

$$
v^2 = \omega^2 (A^2 - y^2)
$$

Once the block detaches, it experiences a constant downward acceleration $g$. The maximum additional height $h_{	ext{free}}$ it can climb is given by the standard kinematic formula:

$$
h_{	ext{free}} = \frac{v^2}{2g}
$$

Substituting our expression for $v^2$ into this formula:

$$
h_{	ext{free}} = \frac{\omega^2(A^2 - y^2)}{2g}
$$

Now, the total height $h$ above the mean position is:

$$
h = y + h_{	ext{free}} = y + \frac{\omega^2(A^2 - y^2)}{2g}
$$

This is our master equation! It expresses the total height $h$ as a quadratic function of the detachment position $y$.

---

### The Calculus of Optimization

To find the value of $y$ that maximizes $h$, we differentiate $h$ with respect to $y$:

$$
\frac{dh}{dy} = \frac{d}{dy} \left( y + \frac{\omega^2 A^2}{2g} - \frac{\omega^2 y^2}{2g} ight)
$$

$$
\frac{dh}{dy} = 1 - \frac{\omega^2 y}{g}
$$

Setting this derivative to zero to find the critical point:

$$
1 - \frac{\omega^2 y}{g} = 0 \implies y = \frac{g}{\omega^2}
$$

---

### Verifying the Maximum

To ensure that this critical point corresponds to a maximum, we perform the second derivative test:

$$
\frac{d^2h}{dy^2} = -\frac{\omega^2}{g}
$$

Since both $\omega^2$ and $g$ are positive physical constants, the second derivative is strictly negative:

$$
\frac{d^2h}{dy^2} < 0
$$

This mathematically guarantees that $y = \frac{g}{\omega^2}$ yields the **maximum** total height.

---

### The Physical Reality and Constraints

For this solution to be physically meaningful, the detachment point must lie within the range of the SHM oscillation. That is, the detachment displacement $y$ cannot exceed the amplitude $A$:

$$
y < A \implies \frac{g}{\omega^2} < A \implies A\omega^2 > g
$$

This is precisely the constraint given in the problem! If $A\omega^2 \le g$, the block would never reach the optimal detachment point, and the maximum height would simply be achieved by letting the block reach its natural SHM peak at $y = A$.

Thus, the optimal distance from the mean position is:

$$
\mathbf{y = \frac{g}{\omega^2}}
$$

Visualized Solution

Understanding the Physical Setup

Velocity at Distance $y$

Free Fall Under Gravity

Formulating Total Height $h$

Condition for Maximum Height

Differentiating $h$ with respect to $y$

Finding the Critical Point

Verifying the Maximum

Checking the Physical Constraint

Conclusion

The Sigma Insight: Force and Energy Method in SHM

Introduction

Analyzing the Setup

The Master Equation

The Calculus of Optimization

Verifying the Maximum

The Physical Reality and Constraints

Similar Questions

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Visualized Solution

Understanding the Physical Setup

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

Free Fall Under Gravity

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

Condition for Maximum Height

Differentiating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h with respect to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Finding the Critical Point

Verifying the Maximum

Checking the Physical Constraint

Conclusion

The Sigma Insight: Force and Energy Method in SHM

Introduction

Analyzing the Setup

The Master Equation

The Calculus of Optimization

Verifying the Maximum

The Physical Reality and Constraints

Similar Questions

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