In linear simple harmonic motion of a particle at mean position:
Velocity is maximum and acceleration is minimum
Understanding Simple Harmonic Motion at the Mean Position
In linear Simple Harmonic Motion (SHM), a particle oscillates back and forth about a fixed point known as the mean position or equilibrium position. This position is where the net force acting on the particle is zero.
Velocity of Particle at the Mean Position
The velocity \(v\) of a particle undergoing linear SHM at a displacement \(x\) from the mean position is given by the equation:
\[ v = \omega \sqrt{A^2 - x^2} \]where \(\omega\) is the angular frequency and \(A\) is the amplitude of the motion. At the mean position, the displacement \(x\) is zero, i.e., \(x=0\).
Substituting \(x=0\) into the velocity equation, we get the velocity at the mean position:
\[ v_{\text{mean}} = \omega \sqrt{A^2 - 0^2} = \omega \sqrt{A^2} = \omega A \]Since \(\omega\) and \(A\) are constants for a given SHM, the magnitude of the velocity at the mean position, \( \omega A \), is the maximum possible speed of the particle during its oscillation.
Acceleration of Particle at the Mean Position
The acceleration \(a\) of a particle undergoing linear SHM at a displacement \(x\) from the mean position is given by the equation:
\[ a = -\omega^2 x \]where \(\omega\) is the angular frequency and \(x\) is the displacement from the mean position. The negative sign indicates that the acceleration is always directed towards the mean position.
At the mean position, the displacement \(x\) is zero, i.e., \(x=0\). Substituting \(x=0\) into the acceleration equation, we get the acceleration at the mean position:
\[ a_{\text{mean}} = -\omega^2 (0) = 0 \]The acceleration at the mean position is zero. The magnitude of acceleration is directly proportional to the displacement from the mean position, so it is minimum (zero) when the displacement is zero at the mean position.
Summary of Velocity and Acceleration at Mean Position
Based on the formulas for velocity and acceleration in SHM, we find:
- At the mean position (\(x=0\)), the magnitude of velocity is \( \omega A \), which is its maximum value.
- At the mean position (\(x=0\)), the acceleration is \( 0 \), which is its minimum magnitude.
Other points in the motion exhibit different values:
- At the extreme positions (\(x = \pm A\)), the velocity is \( v = \omega \sqrt{A^2 - (\pm A)^2} = 0 \), which is minimum.
- At the extreme positions (\(x = \pm A\)), the acceleration is \( a = -\omega^2 (\pm A) = \mp \omega^2 A \). The magnitude \( \omega^2 A \) is maximum.
Analyzing the Options for SHM at Mean Position
Let's examine the given options in light of our understanding:
- Option 1: Velocity is maximum and acceleration is minimum. This statement correctly describes the conditions at the mean position in linear SHM.
- Option 2: Velocity is minimum and acceleration is maximum. This describes the conditions at the extreme positions, not the mean position.
- Option 3: both velocity and acceleration are minimum. Velocity is maximum at the mean position, so this is incorrect.
- Option 4: both velocity and acceleration are maximum. Velocity is maximum but acceleration is minimum (zero) at the mean position, so this is incorrect.
Therefore, at the mean position in linear simple harmonic motion, the velocity is at its maximum magnitude, and the acceleration is at its minimum magnitude (which is zero).
Important Questions from Simple Harmonic Motion
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
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