A particle is executing simple harmonic motion with time period 2 s and amplitude 1 cm . If D and d are the total distance and displacement covered by the particle in 12.5 s , then D/d is

Solution

Physics sir · 4 steps

Break 12.5 s into whole periods plus a quarter: whole periods add distance but leave no displacement.

1. Count the periods. With T = 2 s, the time 12.5 s is 12.5/2 = 6.25 periods. That is 6 full oscillations plus T/4.
2. Distance travelled. In one full oscillation the particle covers 4A = 4 cm, so 6 oscillations give 24 cm. In a quarter period it goes from the mean position to an extreme, or from an extreme to the mean, which is A = 1 cm.

   D = 6(4A) + A = 24 + 1 = 25 cm

3. Displacement. After 6 full oscillations the particle is back where it started, so those contribute zero displacement. Only the last quarter period counts, and it moves the particle between the mean and an extreme, so d = A = 1 cm. This holds whether it starts at the mean position or at an extreme.
4. So. D/d = (25 cm)/(1 cm) = 25, which is option (c).

The idea to keep. In SHM, every full period adds 4A to the distance and nothing to the displacement. Split the time into whole periods plus the leftover, then handle only the leftover carefully.

Answer from the NTA final answer key. Solution by PrepWiser.