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NCERT Exemplar · Q24

Q.What is the ratio between the distance travelled by the oscillator in one time period and amplitude?

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An oscillator completes one full cycle in one time period, traveling from equilibrium to maximum displacement and back twice. The total distance is four times the amplitude, giving a ratio of 4:1.

Why distance differs from displacement in SHM

In simple harmonic motion, we must distinguish carefully between displacement (a vector quantity that can be positive or negative) and distance (the actual path length traveled, always positive). Over one complete time period TT, the oscillator returns to its starting point, so the net displacement is zero. But the particle has certainly moved—it has traced out a specific path whose length we can measure.

The key insight is to track the oscillator's journey segment by segment through one complete cycle.

Tracing the path through one period

Suppose the oscillator starts at the mean (equilibrium) position and moves with amplitude AA.

  1. First quarter-period (0→T/40 \to T/4): The oscillator moves from the mean position to the positive extreme position +A+A. Distance traveled = AA.

  2. Second quarter-period (T/4→T/2T/4 \to T/2): It reverses direction and returns from +A+A back to the mean position. Distance traveled = AA.

  3. Third quarter-period (T/2→3T/4T/2 \to 3T/4): Continuing through the mean position, it travels to the negative extreme position −A-A. Distance traveled = AA.

  4. Fourth quarter-period (3T/4→T3T/4 \to T): Finally, it returns from −A-A back to the mean position, completing the cycle. Distance traveled = AA.

The total distance covered in one complete period is:

Total distance=A+A+A+A=4A\text{Total distance} = A + A + A + A = 4A …

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