Animated Solution for Physics - Oscillations: Time period of a simple pendulum is $T$. The time taken to complete $\frac{5}{8}$ oscillations starting from mean position is $\frac{\alpha }{\beta }T$. The value of $\alpha$ is ......... .

The beauty of Simple Harmonic Motion (SHM) lies in its symmetry and predictability. When a problem asks about a "fraction" of an oscillation, it is testing your fundamental understanding of how distance and time map onto the sine wave of SHM. Let's dive into this fascinating problem!

First, we need to understand what exactly constitutes one complete oscillation. Imagine a pendulum bob resting at its mean position ($x=0$). When released, it travels to the right extreme ($+A$), returns to the mean ($0$), swings to the left extreme ($-A$), and finally comes back to the mean ($0$).

If we add up the distances for each of these four segments, we get:

The problem asks for the time taken to complete $\frac{5}{8}$ of an oscillation. Let's convert this fraction into a physical distance:

This means our particle needs to travel a total distance of $2.5A$.

The Easy Half

To make things simpler, let's break this $2.5A$ journey into two distinct parts: a $2A$ segment and a $0.5A$ segment.

The first $2A$ is exactly half of a full oscillation. The particle goes from the mean to the right extreme and comes back to the mean. Since a full oscillation takes time $T$, this half-journey will take exactly half the time:

The Tricky 0.5A

Now, the particle is back at the mean position, but it still needs to cover the remaining $0.5A$ distance. It will do this by moving towards the left extreme.

Here is where many students make a classic mistake. They assume that since covering $A$ takes $T/4$, covering $A/2$ must take $T/8$. This is incorrect! The velocity of a particle in SHM is not constant; it is maximum at the mean position and zero at the extremes. Therefore, it covers the first half of the amplitude much faster than the second half.

To find the exact time, we use the standard equation of motion for a particle starting from the mean position:

We need the time $t_2$ when the magnitude of displacement is $A/2$:

The smallest angle that satisfies this is $30^{\circ }$, or $\pi /6$ radians.

Since the angular frequency $\omega =\frac{2\pi }{T}$, we can substitute this in:

We now have the times for both legs of the journey. To find the total time, we simply add them up:

To add these fractions, we take a common denominator of 12:

The problem states that this total time is equal to $\frac{\alpha }{\beta }T$. By comparing our result with the given expression, we can clearly see that:

And there we have it! By carefully mapping the physical distance to the mathematical equations of SHM, we've arrived at the perfect answer.