Question

Three identical masses are at the three corners of the triangle, connected by massless identical springs (rest length \( l_0 \)) forming an isosceles right angle triangle. If the two sides of equal length (of length \( 2l_0 \)) lie along the positive x-axis and positive y-axis, then the force on the mass that is not at the origin but on the x-axis is given by \( \hat{i} + \hat{j} \) with \( a \) and \( b \).

• A. \( a = 1 \) and \( b = 0 \)
• B. \( a = 0 \) and \( b = 1 \)
• C. \( a = -\sqrt{2} \) and \( b = 1 \)
• D. \( a = -2 \) and \( b = 0 \)
• E. \( a = -2 \) and \( b = 1 \)

Hint:

In problems involving springs, remember to use Hooke’s Law \( F = -k \Delta x \), where \( k \) is the spring constant and \( \Delta x \) is the displacement.

Answer 1

Answer: (E)

Given: Three identical masses form an isosceles right triangle with sides along the positive $x$ and $y$ axes. The springs have rest length $l_0$, and the two equal triangle sides are of length $2l_0$. 

Let the positions of the three masses be:

• At origin: $(0, 0)$ 
• On x-axis: $(2l_0, 0)$
• On y-axis: $(0, 2l_0)$

We analyze the force on the mass at $(2l_0, 0)$: 

- It is connected to the mass at the origin with a spring stretched by $2l_0 - l_0 = l_0$ (since rest length is $l_0$). - Force from this spring is along the negative x-axis: $$ \vec{F}_1 = -k l_0 \hat{i} $$ - It is also connected to the mass at $(0, 2l_0)$. Distance between them is: $$ \sqrt{(2l_0)^2 + (2l_0)^2} = \sqrt{8l_0^2} = 2\sqrt{2}l_0 $$ So the extension = $2\sqrt{2}l_0 - l_0 = (2\sqrt{2} - 1)l_0$ - Unit vector from $(2l_0, 0)$ to $(0, 2l_0)$ is: $$ \frac{(-2l_0 \hat{i} + 2l_0 \hat{j})}{2\sqrt{2}l_0} = \frac{-\hat{i} + \hat{j}}{\sqrt{2}} $$ - So the force from this spring is: $$ \vec{F}_2 = k(2\sqrt{2} - 1)l_0 \cdot \frac{-\hat{i} + \hat{j}}{\sqrt{2}} $$ Now, add both forces: $$ \vec{F}_{\text{net}} = \vec{F}_1 + \vec{F}_2 $$ Let’s factor and simplify: - Take $k = 1$, $l_0 = 1$ for simplicity. Then: $$ \vec{F}_1 = -1 \hat{i} $$ $$ \vec{F}_2 = (2\sqrt{2} - 1) \cdot \frac{-\hat{i} + \hat{j}}{\sqrt{2}} = (\frac{2\sqrt{2} - 1}{\sqrt{2}})(-\hat{i} + \hat{j}) = (2 - \frac{1}{\sqrt{2}})(-\hat{i} + \hat{j}) $$ Approximate: $\frac{1}{\sqrt{2}} \approx 0.707$ So, $2 - 0.707 \approx 1.293$ Now: $$ \vec{F}_2 \approx 1.293 (-\hat{i} + \hat{j}) = -1.293 \hat{i} + 1.293 \hat{j} $$ Then: $$ \vec{F}_{\text{net}} \approx -1 \hat{i} -1.293 \hat{i} + 1.293 \hat{j} = -2.293 \hat{i} + 1.293 \hat{j} $$ This is approximately: $$ \vec{F} = -2\hat{i} + \hat{j} $$ 
Answer: (e) a = -2 and b = 1