Problem source

The points $A,B,C,D$ and $E$ lie on a thin smooth horizontal table and are equally spaced on a circle with centre $O$ and radius $a$. At each of these points there is a small smooth hole in the table. Five elastic strings are threaded through the holes, one end of each beging attached at $O$ under the table and the other end of each being attached to a particle $P$ of mass $m$ on top of the table. Each of the string has natural length $a$ and modulus of elasticity $\lambda.$ If $P$ is displaced from $O$ to any point $F$ on the table and released from rest, show that $P$ moves with simple harmonic
motion of period $T$, where 
\[
T=2\pi\sqrt{\frac{am}{5\lambda}}.
\]
The string $PAO$ is replaced by one of natural length $a$ and modulus $k\lambda.$ $P$ is displaced along $OA$ from its equilibrium position and released. Show that $P$ still moves in a straight line with simple harmonic motion, and, given that the period is $T/2,$ find $k$.

Solution source

\begin{center}
    \begin{tikzpicture}
        \def\r{3};
        \coordinate (O) at (0,0);
        \coordinate (P) at (1,1);

        \coordinate (A) at ({\r*cos(0)},{\r*sin(0)});
        \coordinate (B) at ({\r*cos(72)},{\r*sin(72)});
        \coordinate (C) at ({\r*cos(144)},{\r*sin(144)});
        \coordinate (D) at ({\r*cos(216)},{\r*sin(216)});
        \coordinate (E) at ({\r*cos(288)},{\r*sin(288)});

        \draw[dashed] (O) circle ({\r});

        \filldraw (O) circle (1pt) node[right] {$O$};
        \filldraw (A) circle (1pt) node[right] {$A$};
        \filldraw (B) circle (1pt) node[above] {$B$};
        \filldraw (C) circle (1pt) node[left] {$C$};
        \filldraw (D) circle (1pt) node[left] {$D$};
        \filldraw (E) circle (1pt) node[below] {$E$};
        \filldraw (P) circle (1pt) node[above right] {$P$};

        \draw[dotted] (O) -- (A); \draw (A)-- (P);
        \draw[dotted] (O) -- (B); \draw (B)-- (P);
        \draw[dotted] (O) -- (C); \draw (C)-- (P);
        \draw[dotted] (O) -- (D); \draw (D)-- (P);
        \draw[dotted] (O) -- (E); \draw (E)-- (P);

        \draw[-latex, blue, ultra thick] (P) -- ($(P)!0.3!(A)$) node[right] {$T_a$};
        \draw[-latex, blue, ultra thick] (P) -- ($(P)!0.3!(B)$) node[right] {$T_b$};
        \draw[-latex, blue, ultra thick] (P) -- ($(P)!0.3!(C)$) node[left] {$T_c$};
        \draw[-latex, blue, ultra thick] (P) -- ($(P)!0.3!(D)$) node[left] {$T_d$};
        \draw[-latex, blue, ultra thick] (P) -- ($(P)!0.3!(E)$) node[right] {$T_e$};
    \end{tikzpicture}
\end{center}

The extension of $OAP$ is $|AP|$ and so the tension $T_a = \frac{\lambda}{a} |AP|$.

To simplify calculations, let $A = a, B = a \omega, C = a \omega^2, \cdots$ where $\omega = e^{2 \pi i/5}$ and let $P = z$. then we can calculate the force as:

\begin{align*}
&&\sum_{p}T_p \mathbf{n}_{z \to p} &= \sum_{p} \frac{\lambda}{a} |z-p| \frac{p-z}{|p-z|} \\
&&&= \frac{\lambda}{a} \sum_{p} ( p - z) \\
&&&= -\frac{5\lambda}{a}z
\end{align*}

Therefore the force has magnitude $\frac{5 \lambda}{a} |OP|$ directly towards the origin. Therefore if we set up our coordinate axis such that $OP$ is the $x$ axis, the particle will remain on the $x$ axis and will move under the equation:
\[
m \ddot{x} + \frac{5 \lambda}{a} x = 0
\]

But then we can say that $P$ moves under SHM with period $\displaystyle 2 \pi \sqrt{\frac{am}{5 \lambda}}$ as required.


Now suppose that $PAO$ has been replaced with the string of modulus $k \lambda$ but that $P$ is along $OA$. 

\begin{align*}
F &= \frac{\lambda}{a}\left ( (a \omega - z) +  (a \omega^2 - z)+ (a \omega^3 -z)+ (a \omega^4 - z) + k(a -z) \right) \\
&= \frac{\lambda}{a}(-a - 4z+ka -kz) \\
&= \frac{\lambda}{a}((k-1)a-(k+4)z)
\end{align*}

Notice that if $z$ is real, this expression is also real, so all forces are acting along $OA$. Therefore the particle will remain on the line $OA$.

We can also notice that the particle will move under the differential equation

\[ m \ddot{x} + \frac{(k+4) \lambda}{a}x = \lambda(k-1) \]

Therefore it will move with SHM about a point slightly displaced from the origin. The period will be: $\displaystyle 2 \pi \sqrt{\frac{ma}{(k+4)\lambda}}$ which is equal to $T/2$ if $(k+4) = 20 \Rightarrow k = 16$