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 Dd is
- 10
- 165
- 25
- 154
Answer
(3) 25
Solution
One period covers 4A; 12.5 s=6 14 T.
[figure]
D=6(4A)+A=25A=25 cm
After whole periods the particle is back at the extreme; the last quarter period takes it to O: d=A=1 cm
Dd=25
Answer: (3)