Problem 51 - Master Resource Book in JEE Main Physics

Question

Two springs have force constants \(k_{1}\) and \(k_{2}\). These are extended through the same distance \(x\). If their elastic energies are \(E_{1}\) and \(E_{2}\), then \(\frac{E_{1}}{E_{2}}\) is equal to (a) \(k_{1}: k_{2}\) (b) \(k_{2}: k_{1}\) (c) \(\sqrt{k_{1}}: \sqrt{k_{2}}\) (d) \(k_{1}^{2}: k_{2}^{2}\)

Step-by-Step Solution

Answer

\(\frac{E_1}{E_2} = \frac{k_1}{k_2}\); the answer is (a) \(k_1 : k_2\).

Step 1: Understand the Elastic Potential Energy Formula

Elastic potential energy for a spring stretched a distance \(x\) is given by the formula \( E = \frac{1}{2} k x^2 \), where \(k\) is the spring's force constant.

Step 2: Write Expressions for Each Spring's Energy

Using the energy formula, the elastic potential energy stored in the first spring is \( E_1 = \frac{1}{2} k_1 x^2 \) and in the second spring is \( E_2 = \frac{1}{2} k_2 x^2 \).

Step 3: Set Up the Ratio of Energies

We need to find the ratio \( \frac{E_1}{E_2} \). Substitute the expressions for \(E_1\) and \(E_2\): \[\frac{E_1}{E_2} = \frac{\frac{1}{2} k_1 x^2}{\frac{1}{2} k_2 x^2}.\]

Step 4: Simplify the Ratio

Notice that \(\frac{1}{2} x^2\) is common in both numerator and denominator. Thus they cancel each other out:\[\frac{E_1}{E_2} = \frac{k_1}{k_2}.\]

Step 5: Conclude with the Answer

The ratio \( \frac{E_1}{E_2} \) simplifies to \( \frac{k_1}{k_2} \). This corresponds to option (a) \(k_1 : k_2\).

Key Concepts

Hooke's Law

Hooke's Law is a fundamental principle in physics that describes how springs work. The law states that the force needed to extend or compress a spring by some distance is proportional to that distance. Mathematically, it is expressed as:\[ F = -k x \]- **F** represents the force applied to the spring.- **k** is the spring constant or force constant.- **x** is the displacement from the equilibrium position.The negative sign indicates that the force exerted by the spring is in the opposite direction of displacement. This is why when you compress or stretch a spring, it pushes or pulls back in an attempt to return to its original state.
Understanding Hooke's Law is crucial to solving problems involving elastic potential energy and spring dynamics.

Force constant

The force constant, denoted as **k**, is a unique property of a spring that measures its stiffness. It determines how much force is needed to stretch or compress a spring by a unit distance. The larger the force constant, the stiffer the spring and the more force is required to change its length. - **Unit of measurement**: Newton per meter (N/m). The force constant plays a crucial role in calculating the elastic potential energy of a spring. It indicates how strongly the spring resists deformation.
This value varies from one spring to another and can be found from experiments by measuring the force and displacement, using the formula from Hooke's Law.

Energy ratio

The energy ratio in the context of springs refers to the comparison between the elastic potential energies in two different springs when they are extended through the same distance. This is a useful concept when you want to understand how two similar systems react under identical conditions.For two springs:- **\( E_1 = \frac{1}{2} k_1 x^2 \)** for the first spring.- **\( E_2 = \frac{1}{2} k_2 x^2 \)** for the second spring.The ratio of these energies, \( \frac{E_1}{E_2} \), simplifies to \( \frac{k_1}{k_2} \) since the \( \frac{1}{2} x^2 \) term cancels out. This means the energy ratio depends directly on the ratio of their force constants. Therefore, the stiffness of the spring heavily influences the energy stored. In the original problem,- If **\(k_1 > k_2\)**, then **\(E_1 > E_2\)**.- Conversely, if **\(k_2 > k_1\)**, then **\(E_2 > E_1\)**.

Spring extension

Spring extension refers to how much a spring is stretched or compressed from its original length. This displacement is an important factor in calculating forces and energy in spring systems.- **Displacement (x)**: The length by which a spring is extended (or compressed) from its natural position.Using Hooke’s Law, if the displacement is **x**, and given a force constant **k**, the force needed to achieve this extension can be calculated. The potential energy stored, as described by the formula \( E = \frac{1}{2} k x^2 \), relies heavily on the square of the extension.This squared relationship means that even a small increase in the displacement results in a much larger increase in energy. This principle explains the proportional rise in energy requirements as springs are extended further.