When a mass M is attached to the spring of force constant k , then the spring stretches by . If the mass oscillates with amplitude l , what will be maximum potential energy stored in the spring

Options

A. ` (kl)/2`
B. 2 kl
C. `1/2 Mgl`
D. Mgl

Answer

To solve the problem step by step, we need to find the maximum potential energy stored in the spring when a mass M is attached to it and oscillates with an amplitude L.

### Step 1: Understanding the System
When a mass M is attached to a spring with a spring constant k, the spring stretches by a distance L due to the weight of the mass. This means that the force exerted by the spring (F_spring = k * L) is equal to the weight of the mass (F_weight = M * g), where g is the acceleration due to gravity.

### Step 2: Relating Spring Constant and Stretch
From the equilibrium condition, we can write:
\[ kL = Mg \]
This equation tells us that the spring constant k multiplied by the stretch L is equal to the weight of the mass.

### Step 3: Maximum Potential Energy in the Spring
The maximum potential energy (U_max) stored in a spring during oscillation is given by the formula:
\[ U_{max} = \frac{1}{2} k A^2 \]
where A is the amplitude of oscillation. In this case, the amplitude A is equal to L.

### Step 4: Substituting Amplitude into the Energy Formula
Substituting A = L into the potential energy formula gives:
\[ U_{max} = \frac{1}{2} k L^2 \]

### Step 5: Substituting k from the Equilibrium Condition
Now, we can substitute k using the relationship we found in Step 2:
\[ U_{max} = \frac{1}{2} \left(\frac{Mg}{L}\right) L^2 \]
This simplifies to:
\[ U_{max} = \frac{1}{2} MgL \]

### Final Result
Thus, the maximum potential energy stored in the spring when the mass oscillates with amplitude L is:
\[ U_{max} = \frac{1}{2} MgL \]