Understanding the Problem: Minimum Time in SHM

The question asks for the minimum time a particle undergoing Simple Harmonic Motion (SHM) takes to travel between two specific points located symmetrically on either side of the mean position. We are given the amplitude and the total time period of the SHM.

Given Information:

• Amplitude (A) = 25 cm
• Time Period (T) = 3 seconds
• The particle moves between points at +12.5 cm and -12.5 cm from the mean position.

Calculating Angular Frequency

The angular frequency ($\omega$) of SHM is related to the time period (T) by the formula:

\(\omega = \frac{2\pi}{T}\)

Substituting the given time period:

\(\omega = \frac{2\pi}{3} \text{ rad/sec}\)

Relating Position to Phase in SHM

The displacement (x) of a particle in SHM can be described by the equation:

\(x = A \sin(\omega t + \phi)\)

where \(A\) is the amplitude, \(\omega\) is the angular frequency, \(t\) is time, and \(\phi\) is the initial phase.

We are interested in the time taken to move from \(x_1 = -12.5\) cm to \(x_2 = +12.5\) cm. These positions are \(x = -A/2\) and \(x = +A/2\) since \(12.5 \text{ cm} = 25/2 \text{ cm} = A/2\).

Finding Phase Angles for the Positions

Let's consider the phase angle (\(\theta = \omega t + \phi\)) corresponding to these positions. We can use the relation \(x = A \sin(\theta)\).

For \(x_2 = +12.5 \text{ cm}\):

\(12.5 = 25 \sin(\theta_2)\)

\(\sin(\theta_2) = \frac{12.5}{25} = \frac{1}{2}\)

The principal value for \(\theta_2\) is \(\frac{\pi}{6}\) radians.

For \(x_1 = -12.5 \text{ cm}\):

\(-12.5 = 25 \sin(\theta_1)\)

\(\sin(\theta_1) = \frac{-12.5}{25} = -\frac{1}{2}\)

The principal value for \(\theta_1\) is \(-\frac{\pi}{6}\) radians.

Calculating the Phase Difference and Time

The minimum time to move from \(x_1\) to \(x_2\) corresponds to traversing the phase difference between the corresponding phase angles, provided the particle moves continuously through the intermediate positions (including the mean position). The phase difference (\(\Delta \theta\)) between \(\theta_1\) and \(\theta_2\) is:

\(\Delta \theta = \theta_2 - \theta_1 = \frac{\pi}{6} - \left(-\frac{\pi}{6}\right) = \frac{\pi}{6} + \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}\) radians.

The time taken (\(\Delta t\)) for this change in phase is given by:

\(\Delta t = \frac{\Delta \theta}{\omega}\)

Substitute the values of \(\Delta \theta\) and \(\omega\):

\(\Delta t = \frac{\frac{\pi}{3}}{\frac{2\pi}{3}} = \frac{\pi}{3} \times \frac{3}{2\pi} = \frac{1}{2}\) seconds.

So, the minimum time required for the particle to move between the two points located at \(\pm 12.5\) cm from the mean position is 0.5 seconds.

Alternative Method Using Reference Circle

Consider the projection of uniform circular motion onto a diameter as SHM. The positions \(x = A/2\) and \(x = -A/2\) correspond to angles in the reference circle. For \(x = A \sin(\theta)\):

• \(x = A/2\) corresponds to angle \(\theta = \pi/6\) (or \(5\pi/6\)).
• \(x = -A/2\) corresponds to angle \(\theta = -\pi/6\) (or \(7\pi/6\)).

Moving from \(x = -A/2\) to \(x = +A/2\) through the mean position corresponds to traversing the angle from \(-\pi/6\) to \(\pi/6\) in the reference circle, which is a total angular displacement of \(\frac{\pi}{3}\) radians.

The total angle for one full cycle is \(2\pi\) radians, which takes time \(T\). The time taken for an angular displacement of \(\Delta \theta\) is:

\(\Delta t = \left(\frac{\Delta \theta}{2\pi}\right) \times T\)

Using \(\Delta \theta = \frac{\pi}{3}\) and \(T = 3\) sec:

\(\Delta t = \left(\frac{\frac{\pi}{3}}{2\pi}\right) \times 3 = \left(\frac{\pi}{3} \times \frac{1}{2\pi}\right) \times 3 = \frac{1}{6} \times 3 = \frac{3}{6} = \frac{1}{2}\) seconds.

Both methods yield the same result.

Conclusion

The minimum time required for the particle executing SHM with amplitude 25 cm and time period 3 sec to move between points located at 12.5 cm on either side of the mean position is 0.5 seconds.