If two similar springs each of spring constant `K_1` are joined in series, the new spring constant and time period would be changed by a factor:

Options

A. `1/2,sqrt2`
B. `1/4,sqrt2`
C. `1/4,2sqrt2`
D. `1/2,sqrt2`

Answer

To solve the problem of finding the new spring constant and time period when two similar springs, each with spring constant \( K_1 \), are joined in series, we can follow these steps:

### Step 1: Understanding Springs in Series
When two springs are connected in series, the total extension \( X_{\text{net}} \) is the sum of the extensions of each spring. If \( X_1 \) is the extension of the first spring and \( X_2 \) is the extension of the second spring, then:
\[ X_{\text{net}} = X_1 + X_2 \]

### Step 2: Relating Force and Extension
The force \( F \) applied to the system is the same for both springs. According to Hooke's Law, we have:
\[ F = K_1 X_1 \quad \text{and} \quad F = K_1 X_2 \]
Thus, we can express the extensions as:
\[ X_1 = \frac{F}{K_1} \quad \text{and} \quad X_2 = \frac{F}{K_1} \]

### Step 3: Finding the Total Extension
Substituting the expressions for \( X_1 \) and \( X_2 \) into the equation for \( X_{\text{net}} \):
\[ X_{\text{net}} = \frac{F}{K_1} + \frac{F}{K_1} = \frac{2F}{K_1} \]

### Step 4: Finding the Effective Spring Constant
The effective spring constant \( K_{\text{net}} \) for the series combination can be defined as:
\[ X_{\text{net}} = \frac{F}{K_{\text{net}}} \]
Equating the two expressions for \( X_{\text{net}} \):
\[ \frac{2F}{K_1} = \frac{F}{K_{\text{net}}} \]
Cancelling \( F \) from both sides (assuming \( F \neq 0 \)):
\[ \frac{2}{K_1} = \frac{1}{K_{\text{net}}} \]
Thus, we find:
\[ K_{\text{net}} = \frac{K_1}{2} \]

### Step 5: Finding the Time Period
The time period \( T \) of a spring-mass system is given by:
\[ T = 2\pi \sqrt{\frac{m}{K}} \]
For the initial spring constant \( K_1 \):
\[ T_1 = 2\pi \sqrt{\frac{m}{K_1}} \]
For the new effective spring constant \( K_{\text{net}} = \frac{K_1}{2} \):
\[ T_2 = 2\pi \sqrt{\frac{m}{K_{\text{net}}}} = 2\pi \sqrt{\frac{m}{\frac{K_1}{2}}} = 2\pi \sqrt{\frac{2m}{K_1}} = \sqrt{2} \cdot T_1 \]

### Step 6: Finding the Change Factor
The change factor for the spring constant is:
\[ \text{Factor for } K = \frac{K_{\text{net}}}{K_1} = \frac{\frac{K_1}{2}}{K_1} = \frac{1}{2} \]
The change factor for the time period is:
\[ \text{Factor for } T = \frac{T_2}{T_1} = \sqrt{2} \]

### Final Answer
- The new spring constant changes by a factor of \( \frac{1}{2} \).
- The time period changes by a factor of \( \sqrt{2} \).