Two springs of equal lengths and equal cross sectional area are made of materials whose young's moduli are in the ratio of 3 : 2. They are suspended and loaded with the same mass. When stretched and released, they will oscillate with time period in the ratio of

Options

A. `sqrt(2) : sqrt(3)`
B. `sqrt(3) : sqrt(2)`
C. `3 : 2`
D. `9 : 4`

Answer

To solve the problem of finding the ratio of the time periods of oscillation of two springs made of materials with different Young's moduli, we can follow these steps:

### Step 1: Understand the relationship between Young's modulus and spring constant
The spring constant \( k \) of a spring is related to the Young's modulus \( Y \) of the material, the cross-sectional area \( A \), and the length \( L \) of the spring by the formula:
\[ k = \frac{Y \cdot A}{L} \]
Since both springs have the same length and cross-sectional area, the spring constants can be expressed as:
\[ k_A = \frac{Y_A \cdot A}{L} \quad \text{and} \quad k_B = \frac{Y_B \cdot A}{L} \]

### Step 2: Find the ratio of the spring constants
Given the ratio of Young's moduli \( \frac{Y_A}{Y_B} = \frac{3}{2} \), we can express the spring constants in terms of this ratio:
\[ \frac{k_A}{k_B} = \frac{Y_A}{Y_B} = \frac{3}{2} \]

### Step 3: Relate the time period to the spring constant
The time period \( T \) of oscillation for a mass \( m \) attached to a spring is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Thus, the time periods for the two springs can be expressed as:
\[ T_A = 2\pi \sqrt{\frac{m}{k_A}} \quad \text{and} \quad T_B = 2\pi \sqrt{\frac{m}{k_B}} \]

### Step 4: Find the ratio of the time periods
To find the ratio of the time periods, we take the ratio \( \frac{T_A}{T_B} \):
\[ \frac{T_A}{T_B} = \frac{2\pi \sqrt{\frac{m}{k_A}}}{2\pi \sqrt{\frac{m}{k_B}}} = \sqrt{\frac{k_B}{k_A}} \]
Substituting the ratio of the spring constants:
\[ \frac{T_A}{T_B} = \sqrt{\frac{2}{3}} \]

### Step 5: Finalize the answer
Thus, the ratio of the time periods \( T_A : T_B \) is:
\[ T_A : T_B = \sqrt{\frac{2}{3}} : 1 \]