What time are the minute and hour hands of a clock perpendicular?

Question

At noon the minute and hour hands of a clock coincide. Assuming the hands of the clock move continuously:
a) What is the first time $T_1$ when they are perpendicular?
b) What is the next time $T_2$ when they again coincide?

I believe I have solved this problem, but I would like someone to verify my answer.

I got $T_1$ to be 12:16 P.M. and 21.818181... seconds ($\frac{180}{11}$ minutes after noon).

I got $T_2$ to be 1:05 P.M. and 27.272727... seconds ($\frac{720}{11}$ minutes after noon).

If there are any questions relating to this problem, please feel free to ask them, as I will do my best to answer them.

Answer 1

If the minute hand has made $x$ revolutions around the clock ($60x$ minutes), the hour hand has made $\frac1{12}x$ revolutions; the difference between these two values represents an angle. Both hands start at 0 revolutions.

For (1), the relevant angle is $\frac14$ of the circle: $$x=\frac14+\frac1{12}x$$ Solving, we get $x=\frac3{11}$, which corresponds to a time of $\frac{180}{11}$ minutes after noon.

For (2) we replace $\frac14$ with 1, since the minute hand has lapped the hour hand. Then $$x=1+\frac1{12}x$$ Solving, we get $x=\frac{12}{11}$, i.e. $\frac{720}{11}$ minutes after noon.

Therefore, both your solutions are correct.

Answer 2

At noon the hands coincide. In $15 + \delta$ minutes, the angle between them will be $90^\circ$.

The minute hand moves at $\dfrac{360^\circ}{60\ \text{min}} = 6$ degrees per minute. The hour hand moves at $\dfrac{360^\circ}{12\ \text{hr}} = \dfrac 12$ degrees per minute. So we need to solve

\begin{align} (15+\delta)\left(6 - \frac 12 \right) &= 90 \\ 15 + \delta &= \frac{180}{11} \\ \delta &= \frac{15}{11} \text{min} \end{align}

So $T_1 = 12:15 + 0:01\frac{4}{11} = 12:16\frac{4}{11}$.

The hands of the clock will be $90^\circ$ apart every $1 + \epsilon$ hours after that, and they will return to their original position after $11$ changes.

\begin{align} 11(1 + \epsilon) &= 12 \\ 1 + \epsilon &= \frac{12}{11} \\ \epsilon &= \frac{1}{11} \text{hr} \\ \epsilon &= 5\frac{5}{11} \text{min} \end{align}

So

$T_2 = 12:16\frac{4}{11} + 1:05\frac{5}{11} = 1:05\frac{9}{11}$

$T_3 = 1:05\frac{9}{11} + 1:05\frac{5}{11} = 2:11\frac{3}{11}$

$\vdots$