Physics · Oscillations · NEET
Yes. Acceleration a = -omega^2 x, so at the mean position x = 0 and acceleration is zero. This is the point where the pull-back force is zero but the speed is maximum. So at the mean position: acceleration = 0, velocity = maximum.
Because a = -omega^2 x depends on x. At the extreme position x = A (the amplitude), which is the largest possible displacement. So the magnitude of acceleration is largest here: a(max) = omega^2 A. At the same point velocity is zero, because the particle stops for an instant before turning back.
The negative sign shows direction, not that acceleration is 'negative'. It says acceleration always points opposite to displacement, that is, toward the mean position. When the particle is on the right (x positive), the pull is to the left; when on the left (x negative), the pull is to the right. This restoring nature is the reason the motion is oscillatory.
Velocity is v = omega times root(A^2 - x^2); it is maximum at the mean position and zero at the extremes. Acceleration is a = -omega^2 x; it is zero at the mean position and maximum at the extremes. So they are exactly out of step: where one is largest, the other is smallest. Their phase difference is pi/2 (90 degrees).
Start from displacement x = A sin(omega t). Differentiate twice: velocity v = omega A cos(omega t), then acceleration a = -omega^2 A sin(omega t). Since A sin(omega t) = x, this is exactly a = -omega^2 x. So the time form and the displacement form are the same equation written two ways.
A pendulum hung from the roof of a tall building moves freely to and fro as a simple harmonic oscillator. The acceleration of the bob is 20 m/s^2 at a distance of 5 m from the mean position. The time period of oscillation is
A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity equals the magnitude of its acceleration. Its time period (in seconds) is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Acceleration in SHM is a = -omega^2 x, where omega is the angular frequency and x is the displacement from the mean position. In time form it is a = -omega^2 A sin(omega t).
The maximum acceleration is a(max) = omega^2 A, where A is the amplitude. It occurs at the extreme positions (x = A), where the particle is farthest from the mean position.
Acceleration is zero at the mean position, where x = 0. This is because a = -omega^2 x, so zero displacement gives zero acceleration.
Yes. From a = -omega^2 x, acceleration is directly proportional to displacement in magnitude, and it always points opposite to the displacement (toward the mean position). This proportionality is the defining condition of SHM.
The phase difference is pi radians (180 degrees). Acceleration is always opposite in sign to displacement, so when displacement is maximum positive, acceleration is maximum negative.
