Calculating Time in Simple Harmonic Motion (SHM)

The problem asks for the time taken by a particle executing Simple Harmonic Motion (SHM) to move from its extreme position to half its amplitude. We are given the amplitude and the time period of the SHM.

Understanding the Given Parameters

• Amplitude of SHM, \(A = 9\)
• Time Period of SHM, \(T = 4\) seconds

We need to find the time taken to move from the extreme position (let's say \(x = +A\)) to \(x = +A/2\).

Equation of Motion for SHM

When a particle starts its Simple Harmonic Motion (SHM) from the extreme position, its displacement \(x(t)\) at time \(t\) can be described by the equation:

\(x(t) = A \cos(\omega t)\)

where:

• \(A\) is the amplitude
• \(\omega\) is the angular frequency
• \(t\) is the time elapsed

Calculating Angular Frequency

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

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

Substituting the given time period \(T = 4\) seconds:

\(\omega = \frac{2\pi}{4} = \frac{\pi}{2} \) rad/s

Finding the Time Taken

The particle starts at the extreme position \(x = A\) at \(t = 0\). We want to find the time \(t\) when its position is \(x = A/2\).

Substitute \(x(t) = A/2\) into the equation of motion:

\(\frac{A}{2} = A \cos(\omega t)\)

Divide both sides by \(A\):

\(\frac{1}{2} = \cos(\omega t)\)

We know that \(\cos(\frac{\pi}{3}) = \frac{1}{2}\). Therefore,

\(\omega t = \frac{\pi}{3}\)

Now, substitute the value of \(\omega = \frac{\pi}{2}\) rad/s:

\(\left(\frac{\pi}{2}\right) t = \frac{\pi}{3}\)

Solve for \(t\):

\(t = \frac{(\pi/3)}{(\pi/2)} = \frac{\pi}{3} \times \frac{2}{\pi}\)

\(t = \frac{2}{3}\) seconds

Thus, the time taken by the particle to move from the extreme position to half the amplitude is \(\frac{2}{3}\) seconds.

Conclusion

Based on the calculations using the SHM equation starting from the extreme position, the time required to reach half the amplitude is \(\frac{2}{3}\) seconds.