CONSIDER A SHM MOTION OF TIME PERIOD t CALCULATE THE TIME TAKEN FOR DISPLACEMENT TO CHANGE VALUE FROM HALF THE AMPLITUDE TO THE AMPLITUDE.

To tackle the problem of calculating the time taken for the displacement in simple harmonic motion (SHM) to change from half the amplitude to the full amplitude, we can break it down step by step. Let's start by defining some key concepts and then apply them to find the solution.

Understanding Simple Harmonic Motion

In SHM, an object oscillates back and forth around an equilibrium position. The key parameters involved are:

• A: Amplitude, which is the maximum displacement from the equilibrium position.
• T: Time period, the time taken for one complete cycle of motion.
• t: Time taken for a specific displacement.

Displacement in SHM

The displacement \( x \) at any time \( t \) can be described by the equation:

\( x(t) = A \cos\left(\frac{2\pi}{T} t\right) \)

Here, \( A \) is the amplitude, and \( T \) is the time period of the motion. We want to find the time taken for the displacement to change from \( \frac{A}{2} \) (half the amplitude) to \( A \) (the full amplitude).

Setting Up the Equations

First, let's find the time when the displacement is \( \frac{A}{2} \):

\( \frac{A}{2} = A \cos\left(\frac{2\pi}{T} t_1\right) \)

Dividing both sides by \( A \) gives:

\( \frac{1}{2} = \cos\left(\frac{2\pi}{T} t_1\right) \)

The cosine function equals \( \frac{1}{2} \) at angles of \( \frac{\pi}{3} \) and \( \frac{5\pi}{3} \). Thus, we can write:

\( \frac{2\pi}{T} t_1 = \frac{\pi}{3} \) or \( \frac{2\pi}{T} t_1 = \frac{5\pi}{3} \)

Calculating \( t_1 \)

Solving for \( t_1 \):

\( t_1 = \frac{T}{6} \) or \( t_1 = \frac{5T}{6} \)

Since we are interested in the first instance when the displacement is \( \frac{A}{2} \), we take \( t_1 = \frac{T}{6} \).

Finding the Time for Full Amplitude

Next, we need to determine the time when the displacement is equal to the amplitude \( A \):

\( A = A \cos\left(\frac{2\pi}{T} t_2\right) \)

This simplifies to:

\( 1 = \cos\left(\frac{2\pi}{T} t_2\right) \)

The cosine function equals 1 at \( t_2 = 0 \) and at every integer multiple of the time period \( T \). For our purposes, we can consider the first instance after \( t_1 \), which is at \( t_2 = 0 \) (the start of the cycle).

Calculating the Time Interval

Now, we find the time taken for the displacement to change from \( \frac{A}{2} \) to \( A \):

\( \Delta t = t_2 - t_1 \)

Since \( t_2 \) is the start of the cycle (0) and \( t_1 = \frac{T}{6} \):

\( \Delta t = 0 - \frac{T}{6} = -\frac{T}{6} \)

However, since we are looking for the time taken to go from \( \frac{A}{2} \) to \( A \), we consider the positive interval:

\( \Delta t = \frac{T}{6} \)

Final Result

Thus, the time taken for the displacement to change from half the amplitude to the full amplitude is:

\( \frac{T}{6} \)

This means that in one complete cycle of SHM, it takes one-sixth of the period to move from half the amplitude to the full amplitude. This relationship highlights the periodic nature of SHM and how displacement changes over time.