An object is attached to a vertical unstretched light spring suspended from a ceiling and slowly lowered to its equilibrium position. This stretches the spring by x. If the same object is attaced to the same vertical spring but permitted to fall suddenly, the spring stretched by

Options

A. `x//2`
B. `x`
C. `2x`
D. `4x`

Answer

To solve the problem, we need to analyze two scenarios involving a spring and an object attached to it.

### Step-by-Step Solution:

1. **Understanding the First Scenario (Slowly Lowered):**
- When the object is slowly lowered to its equilibrium position, it stretches the spring by a distance \( x \).
- At this position, the force exerted by the spring (Hooke's Law) is equal to the weight of the object.
- Mathematically, this can be expressed as:
\[ kx = mg \]
- Here, \( k \) is the spring constant, \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity.

2. **Finding the Spring Constant \( k \):**
- Rearranging the equation gives us:
\[ k = \frac{mg}{x} \]

3. **Understanding the Second Scenario (Suddenly Released):**
- In this scenario, the object is attached to the spring and then released suddenly from the same position.
- The object will fall under the influence of gravity and will stretch the spring further than \( x \).
- Let’s denote the additional stretch of the spring when the object falls as \( y \). Therefore, the total stretch of the spring in this case will be \( x + y \).

4. **Applying Energy Conservation:**
- When the object falls, it converts its gravitational potential energy into elastic potential energy of the spring.
- The gravitational potential energy when the object has fallen a distance \( y \) is:
\[ U_g = mg(y + x) \]
- The elastic potential energy stored in the spring when stretched by \( y + x \) is:
\[ U_s = \frac{1}{2} k (y + x)^2 \]

5. **Setting Up the Energy Conservation Equation:**
- At the lowest point, all gravitational potential energy is converted into elastic potential energy:
\[ mg(y + x) = \frac{1}{2} k (y + x)^2 \]

6. **Substituting for \( k \):**
- Substitute \( k = \frac{mg}{x} \) into the energy equation:
\[ mg(y + x) = \frac{1}{2} \left(\frac{mg}{x}\right) (y + x)^2 \]

7. **Canceling \( mg \):**
- We can cancel \( mg \) from both sides (assuming \( m \neq 0 \)):
\[ (y + x) = \frac{1}{2x} (y + x)^2 \]

8. **Rearranging the Equation:**
- Rearranging gives:
\[ 2x(y + x) = (y + x)^2 \]
- This simplifies to:
\[ (y + x)(y + x - 2x) = 0 \]
- Thus, we have:
\[ y + x = 0 \quad \text{or} \quad y = x \]

9. **Finding Total Stretch:**
- Since \( y + x = 0 \) is not a valid solution in this context, we focus on the second part:
\[ y = x \]
- Therefore, the total stretch of the spring when the object falls suddenly is:
\[ y + x = x + 2x = 2x \]

### Final Answer:
The spring stretches by \( 2x \) when the object is allowed to fall suddenly.