A massless spring of length l and spring constant k is placed vertically on a table. A ball of mass m is just kept on top of the spring. The maximum velocity of the ball is

Options

A. `gsqrt((m)/(k))`
B. `gsqrt((2m)/(k))`
C. `g/2sqrt((m)/(k))`
D. `gsqrt((m)/(2k))`

Answer

To find the maximum velocity of a ball of mass \( m \) placed on top of a massless spring with spring constant \( k \), we can follow these steps:

### Step 1: Understand the System
Initially, the spring is at its natural length \( l \). When the ball of mass \( m \) is placed on the spring, it will compress the spring due to the gravitational force acting on it.

### Step 2: Determine the Compression of the Spring
When the ball is placed on the spring, it compresses the spring by a distance \( x_1 \). The gravitational potential energy lost by the ball when it falls a distance \( x_1 \) is given by:
\[ \text{Potential Energy} = m g x_1 \]
This energy is converted into the elastic potential energy stored in the spring, which is given by:
\[ \text{Elastic Potential Energy} = \frac{1}{2} k x_1^2 \]
Setting these two energies equal gives us:
\[ m g x_1 = \frac{1}{2} k x_1^2 \]

### Step 3: Solve for \( x_1 \)
Rearranging the equation:
\[ \frac{1}{2} k x_1^2 - m g x_1 = 0 \]
Factoring out \( x_1 \):
\[ x_1 \left( \frac{1}{2} k x_1 - m g \right) = 0 \]
This gives us two solutions: \( x_1 = 0 \) (the trivial solution) or:
\[ \frac{1}{2} k x_1 = m g \implies x_1 = \frac{2 m g}{k} \]

### Step 4: Find the Equilibrium Position
The equilibrium position \( x_0 \) can be found by setting the spring force equal to the weight of the ball:
\[ k x_0 = m g \implies x_0 = \frac{m g}{k} \]

### Step 5: Calculate the Amplitude of Motion
The amplitude \( A \) of the oscillation is the difference between the maximum compression \( x_1 \) and the equilibrium position \( x_0 \):
\[ A = x_1 - x_0 = \frac{2 m g}{k} - \frac{m g}{k} = \frac{m g}{k} \]

### Step 6: Determine the Angular Frequency
The angular frequency \( \omega \) of the simple harmonic motion is given by:
\[ \omega = \sqrt{\frac{k}{m}} \]

### Step 7: Calculate the Maximum Velocity
The maximum velocity \( v_{max} \) in simple harmonic motion is given by:
\[ v_{max} = A \omega \]
Substituting the expressions for \( A \) and \( \omega \):
\[ v_{max} = \left(\frac{m g}{k}\right) \left(\sqrt{\frac{k}{m}}\right) = \frac{m g}{k} \cdot \sqrt{\frac{k}{m}} = \sqrt{\frac{m g^2}{k}} \]

### Final Result
Thus, the maximum velocity of the ball is:
\[ v_{max} = \sqrt{\frac{m g^2}{k}} \]