A Random Picture

Because of the symmetry of the pendulum's oscillation, we only need to consider a quarter of the period, when $x=A \sin (\omega t)$ with $0 \leq t \leq \frac{\pi}{2\omega}$ .

The time that $\frac{|x|}{A} \leq \epsilon$ (or $\sin (\omega t) \leq \epsilon$ ) is $t=\frac{\arcsin \epsilon}{\omega}$ .

Therefore, the probility that the picture will capture the pendulum in a position such that $\frac{|x|}{A} \leq \epsilon$ is

$p=\frac{\frac{\arcsin \epsilon}{\omega}}{\frac{\pi}{2\omega}}=\frac{2 \arcsin \epsilon}{\pi}=0.0638.$

Just for beginners, one can refer to pendulum motion to understand more, especially the derivation. Here we can presume the equation for angular displacement:

$x = A \sin (\sqrt{\frac{g}{L}} t)$

Again, we know that the period of a pendulum is given by $T = 2π\sqrt{\frac{L}{g}}$ . $(*)$

In fact, to simplify the equations to come, we can sub in $\omega = \sqrt{\frac{L}{g}}$ , the angular frequency.

As with other such questions involving pendulums, notice that a pendulum has nice symmetry, so the probability that we want, is:

$\frac{ \text{amount of time when the inequality in question is satisfied}}{\frac{1}{4} \ \text{of the period}}$ $(**)$

Now, basically, since we "start" the time from $t_{\epsilon} = 0$ , we have that the time when $\frac{|x|}{A} \leq \frac{1}{10}$ , is when $\frac{A \sin (\omega t_{\epsilon})}{A} = \frac{1}{10}$ , in a more polished form, we want $\sin(\omega t_{\epsilon}) = \frac{1}{10}$ , or $t_{\epsilon} = \sin^{-1}( \frac{1}{10} ) \times \omega$ .

Subbing this into $(*)$ , along with $(**)$ , we can get the desired answer as:

$\text{Required probability} = \frac{ \sin^{-1}( \frac{1}{10} ) \times \omega}{\frac{1}{4} \times 2π\omega} = \frac{2 \sin^{-1}(\frac{1}{10})}{π} = 0.0638$

P.S. At first I forgot to keep $\sin^{-1}{\frac{1}{10}}$ in radians and I got a really big answer which was obviously wrong.

Let us say the equation of motion is:

$x = A\sin(\omega t)$

When $\frac{x}{A} = \frac{1}{10} , \sin(\omega t_{1}) =\frac{1}{10}, \Rightarrow t_{1} \approx \frac{0.1}{\omega}$ .

Now, in one fourth of a complete oscillation , total time taken = $t_{2} = \frac{T}{4} =\frac{\pi}{2 \omega} \approx \frac{1.57}{\omega}$ , and time spent under $\frac{|x|}{A} \leq \frac{1}{10}$ is $t_{1}$ .

Hence, required probability = $\frac{t_{1}}{t_{2}} \approx \frac{0.1}{1.57} \approx \boxed{0.0638}$

Let the period of the pendulum be T so we can write the deviation x as a function of time:

$x = A\;sin\left(\frac{2\pi t}{T}\right)$

Solving for t:

$t = \frac{T}{2\pi}\;sin^{-1} \left(\frac{x}{A}\right)$

Differentiating with respect to the position:

$dt = \frac{T}{2\pi \sqrt{A^2-x^2}} \; dx$

The probability to find the pendulum at a position x times the infinitesimal displacement dx is the infinitesimal time interval dt divided by the time in which the movement start to repeat (note that because of the symmetry of the harmonic movement this time is not the period but half of it):

$P(x) dx = \frac{dt}{\frac{T}{2}}$

$P(x) dx =\frac{2}{T} \frac{T}{2\pi \sqrt{A^2-x^2}} \; dx$

Integrating and simplifying:

$P(x) = \frac{1}{\pi \; \sqrt{A^2-x^2}}$

Now to calculate the asked probability we take the sum (or integral) of the probability to find the pendulum in a position such that $-\frac{A}{10} \le x \le \frac{A}{10}$ :

$P = \int_{-\frac{A}{10}}^{\frac{A}{10}} P(x) dx$

$P = \int_{-\frac{A}{10}}^{\frac{A}{10}} \frac{1}{\pi \; \sqrt{A^2-x^2}} dx$

$P = \frac{2 \; sin^{-1}(\frac{1}{10})}{\pi}$

$\boxed{P = 0.0638}$

Assume a point sized body moving on the circumference of a circle of radius $A$ ,

Alt text

The projection of the moving body on the y-axis will execute Simple Harmonic Motion.

Let us assume the point sized body is at $Q$ when the bob of the pendulum will be at distance $\frac{A}{10}$ . Therefore distance $QR$ will be $\frac{A}{10}$ .

The length of the arc $PQ$ corresponding to that angular displacement will be $d = A \times \arcsin (0.1)$ .

We need the probability of finding the bob in between the displacement of $\frac{A}{10}$ & $-\frac{A}{10}$ .

That can be easily found by dividing the time it will stay in the favorable range by the total time period or by dividing the length of the arc of the favorable region by the total circumference.

$\therefore P = \frac{4A \times \arcsin(0.1)}{2A \times \pi}$ $= 0.063$

Since one complete cycle consists of four similar steps as following :

1. Going from 0 to A
2. Coming from A to 0 (just opposite to 1)
3. Going from 0 to -A (same as 1 in reverse direction)
4. Coming from -A to 0 (just opposite to 3)

We can consider any one of the above cases for finding the probability. Let us consider the first case. Let the equation of SHM be $x=Asin \omega t$ . When the bob of the pendulum is at $x= \frac{A}{10}$ , then $\frac{A}{10}=Asin \omega t$ $\implies t= \frac{sin^{-1}0.1}{\omega}$ . For this amount of time in the first case, the bob remains within $\frac{A}{10}$ .

Time taken for $\frac{1}{4}^{th}$ of complete motion= $\frac{T}{4}$ = $\frac{\frac{2 \pi}{\omega}}{4}$ = $\frac{\pi}{2 \omega}$

Thus, the required probability for $\frac{1}{4}^{th}$ of complete motion = $\frac{\frac{sin^{-1}0.1}{\omega}}{\frac{\pi}{2 \omega}}$ =0.0638

You can easily see that multiplying the numerator and denominator by 4, it actually gives the probability for the complete motion. It doesn't make any difference! Hence, calculating for one-fourth of complete cycle we get the required probability for the full cucle.

$\therefore$ Ans. 0.0638