Physics · Oscillations · NEET
Imagine a particle moving at constant speed on a circle of radius A. Drop a straight line (a perpendicular) from the particle to one diameter, say the x-axis. The foot of that line is the projection. As the particle goes round, this foot slides left and right between -A and +A. That sliding motion is exact SHM. So we do not say the SHM particle moves in a circle. We use an imaginary helper particle on a circle, called the reference particle, and its shadow gives us the real SHM.
The projection can only reach as far as the particle itself. When the reference particle is on the x-axis, its shadow sits at distance equal to the radius from the centre. That is the farthest the shadow ever goes. Amplitude is defined as the maximum displacement from the mean position, so amplitude A = radius of the reference circle.
Follow three fixed steps. Step 1: amplitude A = radius of the circle. Step 2: angular frequency w = 2 pi / T, where T is the time for one full revolution. Step 3: check where the particle is at t = 0. If it starts at the maximum on that axis, use cosine; if it starts at the centre (mean position) and moves outward, use sine. Put the sign from the direction of motion. This gives x(t) = A cos(wt + phi).
No. w is the angular speed of the reference particle, measured in radians per second, w = 2 pi / T. The linear speed on the circle is v = A w (radius times angular speed). In SHM we carry over w directly, because the phase angle (wt + phi) is exactly the angle the reference particle has turned through.
phi tells you the starting angle of the reference particle at t = 0. If the particle does not start on the axis, it already has some head-start angle. That head-start is phi. So phi simply records where in the cycle the motion began. The full phase is (wt + phi).
A particle P revolves in a circle. The radius is 3 m, the period of revolution is 4 s, at t = 0 the particle is at the top of the circle, and it revolves as shown. The y-projection of the radius vector of the rotating particle P is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
No. The circle and the reference particle are imaginary helpers. The real SHM object moves only along a straight line. The circle is a tool that makes the maths of SHM easy to see and derive.
Yes. Projecting onto the x-axis gives one SHM, projecting onto the y-axis gives another SHM. Both are valid SHMs with the same A and w, just differing by a phase of 90 degrees (one is sine, the other is cosine).
It is the actual angle the reference particle has turned through, measured from a fixed line. That is the deep reason the SHM phase is an angle in radians.
The speed of the reference particle is A w, and its acceleration (centripetal) is A w squared. Projecting these onto the axis gives velocity and acceleration of the SHM, leading to v = -A w sin(wt) and a = -w squared x.
