Physics · Oscillations · NEET
Amplitude is the maximum distance the particle goes from the mean position (the centre line, usually x = 0). Look at the highest point (crest) of the curve. Its height above the centre line is A. It is also the depth of the lowest point (trough) below the centre line, because SHM is symmetric. Example: if the crest is at x = +1 m and the trough at x = -1 m, then A = 1 m. Do not measure crest-to-trough and call it A; that full swing is 2A.
Time period T is the time for one complete oscillation. On the graph, measure the time gap between two neighbouring crests (or two neighbouring troughs, or two points where the curve crosses the centre line going the SAME way). Example: if a crest is at t = 2 s and the next crest is at t = 10 s, then T = 10 - 2 = 8 s. A common trap: the gap between a crest and the very next trough is only half a period (T/2), not a full T.
Look at where the curve starts, at t = 0. If it starts at the centre line (x = 0) and rises, it is a sine curve: x = A sin(wt). If it starts at the top (x = +A, a crest) at t = 0, it is a cosine curve: x = A cos(wt). If it starts somewhere in between, it has a phase constant phi, so x = A sin(wt + phi). The starting point of the graph directly gives you the phase.
First read A and T from the graph, then w = 2 pi / T. Next read the displacement x at the time asked (just read the curve's height there). Then use a = -w squared times x. The minus sign means acceleration always points back toward the mean position. Example: A = 1 m, T = 8 s gives w = pi/4. At the crest x = +1 m, so a = -(pi/4) squared times 1 = -(pi squared)/16 m/s squared.
The slope of the x-t graph is the velocity. Where the curve is steepest, the particle is moving fastest; this happens as the curve crosses the centre line (mean position), where speed is maximum. Where the curve is flat, at the crest and trough (extreme positions), the slope is zero, so velocity is zero there. So the graph shows: fast at the middle, momentarily stopped at the ends.
Because A is the largest possible displacement in SHM. The restoring force always pulls the particle back to the centre, so it can never travel farther than the amplitude. On the graph this appears as a wave trapped between two horizontal lines at x = +A and x = -A. If a curve went beyond these lines, it would not be SHM.
The x-t graph of a particle performing simple harmonic motion is shown in the figure (amplitude A = 1 m, period T = 8 s, curve at a positive maximum at t = 2 s). The acceleration of the particle at t = 2 s is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Time t is on the horizontal x-axis and displacement x (position from the mean position) is on the vertical y-axis. The curve shows how the particle's position changes as time passes.
It is always a sinusoidal curve, which can be written as x = A sin(wt + phi). Whether it looks like a pure sine or a pure cosine just depends on the phase constant phi, that is, where the particle starts at t = 0.
First find the time period T from the graph (crest to next crest). Then use w = 2 pi / T. You do not read w directly off the graph; you get T first.
Velocity is maximum where the curve is steepest, which is when it crosses the mean position (x = 0). Velocity is zero at the crest and trough (the extreme positions), where the curve momentarily flattens.
Amplitude A is the distance from the centre line to one crest. The peak-to-peak value is crest-to-trough, which equals 2A, twice the amplitude. NEET options often include this 2A value as a trap.
