A particle is executing SHM of amplitude 9 and time period of 4 seconds, then the time taken by it to move from the extreme position to half the amplitude is
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.
Important Questions from Simple Harmonic Motion
- The displacement of a particle is given by $y(t) = K + P \sin^2(\omega t) + Q \sin(\omega t) \cos(\omega t)$. If this represents a simple harmonic motion, the amplitude of its oscillation is:
- Which one of the following equations of motion represents simple harmonic motion?
Assume $A$, $B$, $C$, $D$, $m$, $k$, and $\omega$ are all positive constants. In simple harmonic motion, the particle velocity lags behind the displacement by a phase angle of __________.
- A particle undergoes simple harmonic motion. Determine the phase difference between its instantaneous velocity and instantaneous acceleration.
- A particle executes simple harmonic motion along a straight line. When its displacement from the mean position is $x$, its speed is $v$. If the displacement becomes $2x$, its speed reduces to $v/2$. What is the amplitude of the oscillation?
