Two springs are joined and attached to a mass of 16 kg. The system is then suspended vertically from a rigid support. The spring constant of the two spring are `k_1 and k_2` respectively. The period of vertical oscillations of the system will be

Options

A. `8 pi sqrt((k_1 + k_2)/(k_1k_2))`
B. `pi/2 sqrt((k_1)/(k_2))`
C. `sqrt((k_1+k_2)/(8pi))`
D. `pi sqrt((k_1-k_2)/(2))`

Answer

To solve the problem of finding the period of vertical oscillations of a mass attached to two springs in series, we can follow these steps:

### Step 1: Understand the System
We have a mass \( m = 16 \, \text{kg} \) attached to two springs with spring constants \( k_1 \) and \( k_2 \). The springs are connected in series.

### Step 2: Find the Equivalent Spring Constant
When two springs are connected in series, the equivalent spring constant \( k_{\text{equiv}} \) can be calculated using the formula:
\[ \frac{1}{k_{\text{equiv}}} = \frac{1}{k_1} + \frac{1}{k_2} \]
This can be rearranged to:
\[ k_{\text{equiv}} = \frac{k_1 k_2}{k_1 + k_2} \]

### Step 3: Write the Equation of Motion
When the mass is displaced from its equilibrium position, the forces acting on it are the weight of the mass \( mg \) acting downwards and the spring force \( k_{\text{equiv}} x \) acting upwards. The equation of motion can be written as:
\[ m a = mg - k_{\text{equiv}} x \]
Where \( a \) is the acceleration and \( x \) is the displacement from the equilibrium position.

### Step 4: Apply Newton's Second Law
Using Newton's second law, we can express the acceleration \( a \) as \( \frac{d^2x}{dt^2} \):
\[ m \frac{d^2x}{dt^2} = mg - k_{\text{equiv}} x \]
This simplifies to:
\[ \frac{d^2x}{dt^2} = g - \frac{k_{\text{equiv}}}{m} x \]

### Step 5: Identify the Oscillation Equation
The above equation can be rearranged to resemble the standard form of simple harmonic motion:
\[ \frac{d^2x}{dt^2} + \frac{k_{\text{equiv}}}{m} x = g \]
The term \( \frac{k_{\text{equiv}}}{m} \) represents the angular frequency squared \( \omega^2 \).

### Step 6: Find the Period of Oscillation
The period \( T \) of the oscillation is given by:
\[ T = 2\pi \sqrt{\frac{m}{k_{\text{equiv}}}} \]
Substituting \( k_{\text{equiv}} \) into the equation:
\[ T = 2\pi \sqrt{\frac{m}{\frac{k_1 k_2}{k_1 + k_2}}} \]
This simplifies to:
\[ T = 2\pi \sqrt{\frac{m (k_1 + k_2)}{k_1 k_2}} \]

### Step 7: Substitute the Mass
Now, substituting \( m = 16 \, \text{kg} \):
\[ T = 2\pi \sqrt{\frac{16 (k_1 + k_2)}{k_1 k_2}} = 8\pi \sqrt{\frac{k_1 + k_2}{k_1 k_2}} \]

### Final Result
Thus, the period of vertical oscillations of the system is:
\[ T = 8\pi \sqrt{\frac{k_1 + k_2}{k_1 k_2}} \]