Edit Problem

Year

Paper

Question Number

Course

Section

Worksheet Citation (for copying)

Click the copy button or select the text to copy this citation for use in worksheets.

Problem Text

Solution (Optional)

\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 $(k+4) = 20 \Rightarrow k = 16$

Preview

Problem

Solution

Cancel

Current Ratings

Difficulty Rating: 1700.0

Difficulty Comparisons: 0

Banger Rating: 1500.0

Banger Comparisons: 0