Two springs have their force constants in the ratio of `3:4`. Both the springs are stretched by applying equal force F. If elongation in first spring is x then elogation is second spring is

Options

A. `3x`
B. `4x`
C. `(4)/(3)x`
D. `(3)/(4)x`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between force, spring constant, and elongation
The force \( F \) applied on a spring is related to the spring constant \( k \) and the elongation \( x \) by Hooke's Law, which states:
\[ F = kx \]

### Step 2: Set up the equations for both springs
Let the force constant of the first spring be \( k_1 \) and that of the second spring be \( k_2 \). According to the problem, the ratio of the spring constants is given as:
\[ \frac{k_1}{k_2} = \frac{3}{4} \]
Let the elongation in the first spring be \( x \) and in the second spring be \( x_2 \). Since both springs are stretched by the same force \( F \), we can write:
\[ F = k_1 x \quad \text{(for the first spring)} \]
\[ F = k_2 x_2 \quad \text{(for the second spring)} \]

### Step 3: Equate the forces
Since both expressions equal \( F \), we can set them equal to each other:
\[ k_1 x = k_2 x_2 \]

### Step 4: Solve for \( x_2 \)
Rearranging the equation gives:
\[ x_2 = \frac{k_1}{k_2} x \]

### Step 5: Substitute the ratio of spring constants
Now, substitute the ratio \( \frac{k_1}{k_2} = \frac{3}{4} \) into the equation:
\[ x_2 = \frac{3}{4} x \]

### Conclusion
Thus, the elongation in the second spring \( x_2 \) is:
\[ x_2 = \frac{3}{4} x \]

### Final Answer
The elongation in the second spring is \( \frac{3}{4} x \).

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