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

Answer

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

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

### Step 2: Determine the Equivalent Spring Constant
When 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 Formula for the Period of Oscillation
The period \( T \) of a mass-spring system is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k_{\text{equiv}}}} \]

### Step 4: Substitute the Equivalent Spring Constant
Substituting \( k_{\text{equiv}} \) into the period formula, we have:
\[ T = 2\pi \sqrt{\frac{m}{\frac{k_1 k_2}{k_1 + k_2}}} \]
This simplifies to:
\[ T = 2\pi \sqrt{\frac{m (k_1 + k_2)}{k_1 k_2}} \]

### Step 5: Substitute the Mass Value
Now, substituting the mass \( m = 16 \, \text{kg} \):
\[ T = 2\pi \sqrt{\frac{16 (k_1 + k_2)}{k_1 k_2}} \]

### Step 6: Final Simplification
We can factor out the 16:
\[ T = 4\pi \sqrt{\frac{k_1 + k_2}{k_1 k_2}} \]

### Final Answer
Thus, the period of vertical oscillations of the system is:
\[ T = 4\pi \sqrt{\frac{k_1 + k_2}{k_1 k_2}} \]