Infinite springs with force constant`k , 2k, 4k and 8 k .... respectively are connected in series. The effective force constant of the spring will b

Options

A. 2k
B. k
C. `(k)/(2)`
D. `(k)/(4)`

Answer

To solve the problem of finding the effective force constant of an infinite series of springs with force constants \( k, 2k, 4k, 8k, \ldots \), we will follow these steps:

### Step 1: Understand the Series Connection of Springs
When springs are connected in series, the effective spring constant \( k_{\text{eq}} \) can be found using the formula:
\[ \frac{1}{k_{\text{eq}}} = \frac{1}{k_1} + \frac{1}{k_2} + \frac{1}{k_3} + \ldots \]
In this case, the springs have force constants \( k, 2k, 4k, 8k, \ldots \).

### Step 2: Write Down the Individual Spring Constants
The individual spring constants are:
- \( k_1 = k \)
- \( k_2 = 2k \)
- \( k_3 = 4k \)
- \( k_4 = 8k \)
- and so on.

### Step 3: Set Up the Equation for the Effective Spring Constant
Substituting the values into the formula:
\[ \frac{1}{k_{\text{eq}}} = \frac{1}{k} + \frac{1}{2k} + \frac{1}{4k} + \frac{1}{8k} + \ldots \]

### Step 4: Factor Out \( \frac{1}{k} \)
We can factor out \( \frac{1}{k} \) from the equation:
\[ \frac{1}{k_{\text{eq}}} = \frac{1}{k} \left( 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \right) \]

### Step 5: Identify the Series
The series inside the parentheses is a geometric series where:
- The first term \( a = 1 \)
- The common ratio \( r = \frac{1}{2} \)

### Step 6: Calculate the Sum of the Infinite Series
The sum \( S \) of an infinite geometric series can be calculated using the formula:
\[ S = \frac{a}{1 - r} \]
Substituting the values:
\[ S = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2 \]

### Step 7: Substitute Back to Find \( k_{\text{eq}} \)
Now substituting back into the equation for \( \frac{1}{k_{\text{eq}}} \):
\[ \frac{1}{k_{\text{eq}}} = \frac{1}{k} \cdot 2 \]
Thus,
\[ k_{\text{eq}} = \frac{k}{2} \]

### Conclusion
The effective force constant of the infinite series of springs is:
\[ \boxed{\frac{k}{2}} \]