Question

Three identical masses are at the three corners of the triangle, connected by massless identical springs (rest length l₀) forming an isosceles right angle triangle. If the two sides of equal length (of length 2l₀) lie along 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 ax^ +by^ with

Answer 1

Given that,

Rest length = l₀

Displacement = 2l₀

Force$F = a\hat{x}+b\hat{y}$

If the two sides of equal length lie along positive x-axis and positive y-axis,

We know that,

The restoring force is

$F\propto x$

$F=kx$

According to figure,

AB is 2l₀ and AC is 2l₀.

We need to calculate the value of AC

Using pythagorean theorem

$AC=\sqrt{(AB)^2+(AC)^2}$

Put the value into the formula

$AC=\sqrt{(2l)^2+(2l)^2}$

$AC=2\sqrt{2}l$

We need to calculate the force F₁

Using formula of force

$F_{1}=kx$

$F_{1}=kl$

We need to calculate the force F₂

Using formula of force

$F_{2}=kx_{2}$

Put the value into the formula

$F_{2}=k(2\sqrt{2}l-l)$

$F_{2}=k(2\sqrt{2}-1)$

We need to calculate the force along x axis

Using formula of component of force

$F_{x}=F_{1}+F_{2}\sin45$

$a=kl+k(2\sqrt{2}-1)l\times\dfrac{1}{\sqrt{2}}$

$a=kl\dfrac{3\sqrt{2}-1}{\sqrt{2}}$

$a=kl(3-\dfrac{1}{\sqrt{2}})$

We need to calculate the force along y axis

Using formula of component of force

$F_{y}=F_{2}\sin\theta$

Put the value into the formula

$b=kl\dfrac{1}{\sqrt{2}}$

Hence, The force on the x axis and y axis will be $kl(3-\dfrac{1}{\sqrt{2}})$ and $kl\dfrac{1}{\sqrt{2}}$