Springs of constants k, 2k, 4k, 8k,…….,2048k are connected in series. A mass m is attached to one end and the system is allowed to oscillate. The time period is approximately

Options

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

Explanation

To solve the problem of finding the time period of a mass attached to a series of springs with varying spring constants, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Spring Constants:**
The spring constants given are \( k, 2k, 4k, 8k, \ldots, 2048k \). This is a geometric series where the first term \( a = k \) and the common ratio \( r = 2 \).

2. **Determine the Number of Springs:**
The last spring constant is \( 2048k \). Since \( 2048 = 2^{11} \), the series of spring constants goes from \( k \) (which is \( 2^0 k \)) to \( 2048k \) (which is \( 2^{11} k \)). Therefore, there are \( 12 \) springs in total.

3. **Calculate the Equivalent Spring Constant for Series Connection:**
For springs in series, the equivalent spring constant \( k_{eq} \) is given by:
\[ \frac{1}{k_{eq}} = \frac{1}{k} + \frac{1}{2k} + \frac{1}{4k} + \ldots + \frac{1}{2048k} \]
This can be rewritten as:
\[ \frac{1}{k_{eq}} = \frac{1}{k} \left( 1 + \frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{2048} \right) \]

4. **Sum the Series:**
The series \( 1 + \frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{2048} \) is a geometric series with:
- First term \( a = 1 \)
- Common ratio \( r = \frac{1}{2} \)
- Number of terms \( n = 12 \)

The sum \( S_n \) of a geometric series is given by:
\[ S_n = a \frac{1 - r^n}{1 - r} \]
Substituting the values:
\[ S_{12} = 1 \cdot \frac{1 - \left(\frac{1}{2}\right)^{12}}{1 - \frac{1}{2}} = \frac{1 - \frac{1}{4096}}{\frac{1}{2}} = 2 \left(1 - \frac{1}{4096}\right) \approx 2 \]
(Neglecting the small term \( \frac{1}{4096} \))

5. **Substituting Back to Find \( k_{eq} \):**
Now substituting back:
\[ \frac{1}{k_{eq}} = \frac{1}{k} \cdot 2 \implies k_{eq} = \frac{k}{2} \]

6. **Calculate the Time Period:**
The time period \( T \) of a mass-spring system is given by:
\[ T = 2\pi \sqrt{\frac{m}{k_{eq}}} \]
Substituting \( k_{eq} \):
\[ T = 2\pi \sqrt{\frac{m}{\frac{k}{2}}} = 2\pi \sqrt{\frac{2m}{k}} = 2\sqrt{2\pi} \sqrt{\frac{m}{k}} \]

### Final Answer:
The time period \( T \) is approximately \( 2\sqrt{2\pi} \sqrt{\frac{m}{k}} \).

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