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

To find the ratio of the energies of the springs \( \left( \dfrac{E_1}{E_2} \right) \), we start by analyzing the potential energy stored in a spring, which is given by the formula:

\(E = \dfrac{1}{2} k x^2\)

where \( k \) is the spring constant, and \( x \) is the extension or compression of the spring.

Since the systems are in equilibrium, the force due to the spring must balance the gravitational force on the mass. Thus, for the first spring:

\(k_1 x_1 = m_1 g\)

For the second spring:

\(k_2 x_2 = m_2 g\)

From these, we can solve for \( x_1 \) and \( x_2 \):

\(x_1 = \dfrac{m_1 g}{k_1}\)

\(x_2 = \dfrac{m_2 g}{k_2}\)

The energy stored in the first spring is:

\(E_1 = \dfrac{1}{2} k_1 x_1^2 = \dfrac{1}{2} k_1 \left(\dfrac{m_1 g}{k_1}\right)^2 = \dfrac{1}{2} \dfrac{m_1^2 g^2}{k_1}\)

The energy stored in the second spring is:

\(E_2 = \dfrac{1}{2} k_2 x_2^2 = \dfrac{1}{2} k_2 \left(\dfrac{m_2 g}{k_2}\right)^2 = \dfrac{1}{2} \dfrac{m_2^2 g^2}{k_2}\)

Now, find the ratio \( \dfrac{E_1}{E_2} \):

\(\dfrac{E_1}{E_2} = \dfrac{\dfrac{1}{2} \dfrac{m_1^2 g^2}{k_1}}{\dfrac{1}{2} \dfrac{m_2^2 g^2}{k_2}} = \dfrac{m_1^2 k_2}{m_2^2 k_1}\)

Substitute \( \dfrac{m_1}{m_2} = \alpha \) and \( \dfrac{k_1}{k_2} = \beta \):

\(\dfrac{E_1}{E_2} = \dfrac{\alpha^2 k_2}{k_1} = \dfrac{\alpha^2}{\beta}\)

Thus, the ratio of the energies of the springs is:

\(\dfrac{E_1}{E_2} = \dfrac{\alpha^2}{\beta}\)