Question

A massless spring gets elongated by amount \( x_1 \) under a tension of 5 N. Its elongation is \( x_2 \) under the tension of 7 N. For the elongation of \( 5x_1 - 2x_2 \), the tension in the spring will be:

• A. 15 N
• B. 20 N
• C. 11 N
• D. 39 N

Hint:

The elongation of a spring is directly proportional to the applied force, so use the proportionality constant to find the tension for any given elongation.

Answer 1

The Correct Option is C

To find the tension in the spring for the given elongation of \(5x_1 - 2x_2\), let's use Hooke's Law, which states that the tension or force applied to a spring is directly proportional to its elongation, i.e., \(F = kx\), where \(k\) is the spring constant. 

1. Under a tension of 5 N, the elongation is \(x_1\). Therefore, from Hooke's Law, we have: \(5 = k x_1\)
2. Under a tension of 7 N, the elongation is \(x_2\). Therefore: \(7 = k x_2\)
3. We need to find the tension for an elongation of \(5x_1 - 2x_2\). First, let's express this elongation in terms of the known values: \(T = k (5x_1 - 2x_2)\)
4. Substitute the expressions for \(x_1\) and \(x_2\) from the above equations:
   • \(x_1 = \frac{5}{k}\)
   • \(x_2 = \frac{7}{k}\)
5. Now substitute these back into the formula for \(T\): 

\[T = k \left(5 \cdot \frac{5}{k} - 2 \cdot \frac{7}{k}\right) = k \left(\frac{25}{k} - \frac{14}{k}\right) = k \cdot \frac{11}{k} = 11\]

Therefore, the tension in the spring for the elongation of \(5x_1 - 2x_2\) is 11 N. The correct answer is 11 N.

Answer 2

The elongation of a spring is directly proportional to the applied tension (according to Hooke's law), so we can write: \[ x_1 = k \cdot 5, \quad x_2 = k \cdot 7, \] where \( k \) is the spring constant. Now, for the elongation \( 5x_1 - 2x_2 \), we have: \[ 5x_1 - 2x_2 = 5(k \cdot 5) - 2(k \cdot 7) = k(25 - 14) = k \cdot 11. \] The tension required for this elongation is \( T = k \cdot 11 \), so the tension is 11 N.