Question

There are two spring–block systems as shown. They are in equilibrium. If $\dfrac{m_1}{m_2}=\alpha$ and $\dfrac{k_1}{k_2}=\beta$, then the ratio of the energies of the springs $\left(\dfrac{E_1}{E_2}\right)$ is:

• A. $\dfrac{E_1}{E_2}=\dfrac{\alpha^2}{\beta}$
• B. $\dfrac{E_1}{E_2}=\dfrac{\alpha}{\beta}$
• C. $\dfrac{E_1}{E_2}=\dfrac{\alpha}{\beta^2}$
• D. $\dfrac{E_1}{E_2}=\dfrac{\alpha^2}{\beta^2}$

Hint:

In equilibrium spring problems, first find extension using $kx=mg$. Substitute this extension directly into $E=\tfrac{1}{2}kx^2$ to avoid extra steps.

Answer 1

The Correct Option is A

Concept: For a spring–mass system in equilibrium: \[ kx = mg \] The elastic potential energy stored in a spring is: \[ E = \frac{1}{2}kx^2 \]
Step 1: At equilibrium for the first system: \[ k_1 x_1 = m_1 g \quad \Rightarrow \quad x_1 = \frac{m_1 g}{k_1} \] Similarly, for the second system: \[ k_2 x_2 = m_2 g \quad \Rightarrow \quad x_2 = \frac{m_2 g}{k_2} \]
Step 2: Energy stored in the first spring: \[ E_1 = \frac{1}{2}k_1 x_1^2 = \frac{1}{2}k_1\left(\frac{m_1 g}{k_1}\right)^2 = \frac{m_1^2 g^2}{2k_1} \] Energy stored in the second spring: \[ E_2 = \frac{m_2^2 g^2}{2k_2} \]
Step 3: Take the ratio: \[ \frac{E_1}{E_2}=\frac{m_1^2}{m_2^2}\cdot\frac{k_2}{k_1} \] Step 4: Substitute the given ratios: \[ \frac{m_1}{m_2}=\alpha, \quad \frac{k_1}{k_2}=\beta \Rightarrow \frac{k_2}{k_1}=\frac{1}{\beta} \] \[ \frac{E_1}{E_2}=\alpha^2 \cdot \frac{1}{\beta} \]
Step 5: Hence, \[ \boxed{\frac{E_1}{E_2}=\frac{\alpha^2}{\beta}} \]