Two light springs of same length and same area of cross section are made of materials whose young's modulii of elasticity are in the ratio 1 : 2. When loads of same mass are attached to the two springs, the ratio of their respectively time periods of oscillations is

Options

A. `sqrt(2) : 1`
B. `1 : sqrt(2)`
C. `1 : 2`
D. `2 : 1`

Explanation

To solve the problem, we need to find the ratio of the time periods of oscillation of two springs made from materials with different Young's moduli. Here are the steps to derive the solution:

### Step 1: Understand the relationship between Young's modulus and spring constant
Young's modulus (Y) is defined as the ratio of stress to strain. For a spring, the spring constant (k) can be expressed in terms of Young's modulus as follows:

\[ k = \frac{A \cdot Y}{L} \]

where:
- \( A \) = area of cross-section of the spring
- \( Y \) = Young's modulus of the material
- \( L \) = original length of the spring

### Step 2: Establish the relationship between time period and spring constant
The time period (T) of a mass-spring system is given by:

\[ T = 2\pi \sqrt{\frac{m}{k}} \]

where:
- \( m \) = mass attached to the spring
- \( k \) = spring constant

### Step 3: Find the ratio of the spring constants
Given that the Young's moduli of the two springs are in the ratio \( Y_1 : Y_2 = 1 : 2 \), we can express this as:

\[ Y_1 = Y \quad \text{and} \quad Y_2 = 2Y \]

Using the formula for spring constant, we can write:

\[ k_1 = \frac{A \cdot Y_1}{L} = \frac{A \cdot Y}{L} \]
\[ k_2 = \frac{A \cdot Y_2}{L} = \frac{A \cdot 2Y}{L} = 2 \cdot \frac{A \cdot Y}{L} = 2k_1 \]

Thus, we have:

\[ k_2 = 2k_1 \]

### Step 4: Find the ratio of the time periods
Now, substituting the spring constants into the time period formula, we have:

\[ T_1 = 2\pi \sqrt{\frac{m}{k_1}} \]
\[ T_2 = 2\pi \sqrt{\frac{m}{k_2}} = 2\pi \sqrt{\frac{m}{2k_1}} \]

Now, we can find the ratio of the time periods:

\[ \frac{T_1}{T_2} = \frac{2\pi \sqrt{\frac{m}{k_1}}}{2\pi \sqrt{\frac{m}{2k_1}}} \]

This simplifies to:

\[ \frac{T_1}{T_2} = \frac{\sqrt{\frac{m}{k_1}}}{\sqrt{\frac{m}{2k_1}}} = \frac{\sqrt{2}}{1} = \sqrt{2} \]

### Step 5: Final ratio of time periods
Thus, the ratio of the time periods of the two springs is:

\[ T_1 : T_2 = \sqrt{2} : 1 \]

### Conclusion
The final answer is that the ratio of their respective time periods of oscillation is \( \sqrt{2} : 1 \).

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