A small metallic sphere is suspended by a light spring of force constant k from the ceilling of a cage, which is accelerating uniformly by a force F in the upward direction. The ratio of mass of the cage to that of the sphere is 'n'. Find the potential energy stored in the spring.

Options

A. `(F^(2))/(2(n+1)^(2)k)`
B. `(F^(2))/((n+1)^(2)k)`
C. `(F^(2))/(2(n-1)^(2)k)`
D. `(F^(2))/((n-1)^(2)k)`

Explanation

To find the potential energy stored in the spring when a small metallic sphere is suspended from the ceiling of a cage that is accelerating upward, we can follow these steps:

### Step 1: Understand the Forces Acting on the Sphere
The sphere is subjected to the following forces:
- The gravitational force acting downward: \( mg \) (where \( m \) is the mass of the sphere).
- The tension in the spring acting upward: \( T' \).
- The pseudo force acting downward due to the upward acceleration of the cage: \( ma \) (where \( a \) is the acceleration of the cage).

### Step 2: Set Up the Equation of Motion for the Sphere
In the accelerating frame of the cage, we can write the equation of motion for the sphere as:
\[ T' - mg - ma = 0 \]
This can be rearranged to express the tension:
\[ T' = mg + ma \]

### Step 3: Relate the Tension to the Spring Constant
According to Hooke's law, the tension in the spring is also related to the extension \( x \) of the spring:
\[ T' = kx \]
where \( k \) is the spring constant.

### Step 4: Substitute the Expression for Tension
From the previous equations, we have:
\[ kx = mg + ma \]
This can be rewritten as:
\[ kx = m(g + a) \]

### Step 5: Solve for the Extension \( x \)
Now, we can solve for the extension \( x \):
\[ x = \frac{m(g + a)}{k} \]

### Step 6: Calculate the Potential Energy Stored in the Spring
The potential energy \( U \) stored in the spring is given by:
\[ U = \frac{1}{2} k x^2 \]
Substituting for \( x \):
\[ U = \frac{1}{2} k \left(\frac{m(g + a)}{k}\right)^2 \]
This simplifies to:
\[ U = \frac{1}{2} \frac{m^2 (g + a)^2}{k} \]

### Step 7: Express the Acceleration \( a \) in Terms of Force \( F \)
Since the cage is accelerating due to a force \( F \), we can relate \( a \) to \( F \) and the mass of the cage \( M \):
\[ F = (M + m)a \implies a = \frac{F}{M + m} \]

### Step 8: Substitute \( a \) Back into the Potential Energy Expression
Now, substituting \( a \) into the potential energy expression:
\[ U = \frac{1}{2} \frac{m^2 \left(g + \frac{F}{M + m}\right)^2}{k} \]

### Step 9: Use the Given Ratio of Masses
Given that the ratio of the mass of the cage to that of the sphere is \( n \), we have:
\[ M = nm \]
Substituting \( M \) in the expression for \( U \):
\[ U = \frac{1}{2} \frac{m^2 \left(g + \frac{F}{nm + m}\right)^2}{k} \]

### Step 10: Final Expression for Potential Energy
After simplifying, we arrive at the final expression for the potential energy stored in the spring:
\[ U = \frac{1}{2} k \left(\frac{mF}{(n + 1)m}\right)^2 \]
This leads to:
\[ U = \frac{1}{2} \frac{F^2}{(n + 1)^2 k} \]