A particle executes simple harmonic motion with amplitude $A$ and time period $T$. If the particle starts its motion from one of its extreme positions, what is the total distance covered by the particle in the first $\frac{T}{6}$ of its motion?
$\frac{A}{2}$
Understanding Simple Harmonic Motion and Key Parameters
The question asks about the total distance covered by a particle undergoing Simple Harmonic Motion (SHM) during a specific fraction of its time period. We are given the Amplitude ($A$) and the Time Period ($T$) of the motion. Crucially, the particle starts its journey from one of its extreme positions. We need to find the distance traveled in the initial $\frac{T}{6}$ time.
Analyzing the Motion Parameters in SHM
In Simple Harmonic Motion, a particle oscillates back and forth around a central equilibrium position.
- Amplitude ($A$): This is the maximum displacement of the particle from its mean (equilibrium) position.
- Time Period ($T$): This is the time taken for one complete oscillation (back and forth).
- Starting Point: The particle begins at an extreme position (where displacement is maximum, either $+A$ or $-A$). Let's assume it starts at $x = +A$ at time $t=0$.
- Time Interval: We are interested in the motion during the first $t = \frac{T}{6}$.
Calculating Particle Position Over Time in SHM
The standard equation for the position $x(t)$ of a particle in SHM, when it starts from an extreme position ($x=A$) at $t=0$, is given by:
$x(t) = A \cos(\omega t)$
Here, $\omega$ is the angular frequency, which is related to the time period $T$ by the formula:
$\omega = \frac{2\pi}{T}$
Substituting this into the position equation, we get:
$x(t) = A \cos\left(\frac{2\pi}{T} t\right)$
Determining the Total Distance Covered
We need to find the position of the particle at the beginning ($t=0$) and at the end of the interval ($t=\frac{T}{6}$).
- Position at $t=0$: As given, the particle starts at an extreme position.
$x(0) = A \cos\left(\frac{2\pi}{T} \times 0\right) = A \cos(0) = A \times 1 = A$
- Position at $t=\frac{T}{6}$: Let's calculate the position at the end of the interval.
$x\left(\frac{T}{6}\right) = A \cos\left(\frac{2\pi}{T} \times \frac{T}{6}\right)$
Simplifying the argument of the cosine function:
$\frac{2\pi}{T} \times \frac{T}{6} = \frac{2\pi}{6} = \frac{\pi}{3}$
So, the position is:
$x\left(\frac{T}{6}\right) = A \cos\left(\frac{\pi}{3}\right)$
Since $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$, we have:
$x\left(\frac{T}{6}\right) = A \times \frac{1}{2} = \frac{A}{2}$
Calculating Distance from Position Change
The total distance covered is the total path length traveled. Since the particle starts at $x=A$ (an extreme) and moves towards the mean position ($x=0$) during this time interval ($0$ to $\frac{T}{6}$), the path is monotonic. Therefore, the distance covered is simply the magnitude of the displacement during this interval.
Distance = |Final Position - Initial Position|
Distance $= \left|x\left(\frac{T}{6}\right) - x(0)\right|$
Distance $= \left|\frac{A}{2} - A\right| = \left|-\frac{A}{2}\right|$
Distance $= \frac{A}{2}$
Final Answer Justification
The calculation shows that the total distance covered by the particle in the first $\frac{T}{6}$ of its motion, starting from an extreme position, is $\frac{A}{2}$. This corresponds to one of the given options.
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?
