I've been doing a few old Applied Mathematics exams from the 70s and would like some verification that my solution is correct. Here is the problem:
A particle of mass $m$ is attached to two elastic strings, each of natural length $a$. The other ends of the strings are fixed at points $A$ and $B$ on a horizontal table, where $AB = 2b (>2a)$, and the particle slides smoothly along the table along $AB$ in such a way that neither string slackens.
If an extension $a$ in either string produces a tension $kmg$ in that string, obtain an equation of motion for the particle, and hence write down the period of oscillation.
[8 marks]
So here is the diagram I created. $x$ denotes the displacement of the mass (at point $P$) in a general position during the motion.
The red dashed line indicated the extension. Using Hooke's Law, where $k_{s}$ is the stiffness of the spring. (I am assuming $k_{s}$ is different to the $k$ given in the problem).
$$ T = k_{s}x $$ $$ k_{s} = \dfrac{T}{x} $$
Given $T = kmg$ when $x = a$.
$$ k_{s} = \dfrac{kmg}{a} $$
From the diagram:
Extension of string $AC$ is $(b - a) + x$.
Extension of string $BD$ is $(b - a) - x$.
Equation of motion for mass $m$ using Newton's Second Law is: $$ T_2 - T_1 = m \ddot x $$ $$ k_{s} x_2 - k_{s} x_1 = m \ddot x $$
Where $x_2$ is the extension produces by $T_2$ and $x_1$ is the extension produces by $T_1$.
$$ \dfrac{kmg}{a} \big[ (b - a) - x \big] - \dfrac{kmg}{a} \big[ (b - a) + x \big] = m \ddot x $$
$$ \dfrac{kmg}{a} \big[ b - a - x - b + a - x \big] = m \ddot x $$
$$ \dfrac{kg}{a} \big[ -2x \big] = \ddot x $$
$$ \ddot x = - \dfrac{2kg}{a} x $$
This is in the form of SHM, $ \ddot x = - n^2 x $, where $n^2 = \dfrac{2kg}{a}$.
Period is given by $Period = \dfrac{2\pi}{n}$.
$$ Period = \dfrac{2\pi}{\sqrt{ \dfrac{2kg}{a}}} $$
$$ Period = \dfrac{2 \pi \sqrt{a}}{\sqrt{2kg}} $$
$$ Period = \pi \sqrt{ \dfrac{2a}{kg} } $$
Well, that's my logic, can someone confirm the method is correct. Thanks in advanced.
It looks correct to me.
You could use only a single variable for $\omega=n$ though.
A nice feature is: $b$ does not show up in the end. This is expected, because of the linearity of Hooke's law.