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NEET PHYSICSEasy

The distance covered by a particle undergoing SHM in one time period is: (amplitude = $A$ )

A

zero

B

$A$

C

$2A$

D

$4A$

Step-by-Step Solution

Understand SHM Motion: In one complete time period ( $T$ ), a particle performing Simple Harmonic Motion starts from a point, goes to one extreme, returns to the mean position, goes to the other extreme, and returns to the start.
Breakdown of Path:

From Mean Position ( $x=0$ ) to Extreme Position ( $x=+A$ ): Distance = $A$
From Extreme Position ( $x=+A$ ) back to Mean Position ( $x=0$ ): Distance = $A$
From Mean Position ( $x=0$ ) to other Extreme Position ( $x=-A$ ): Distance = $A$
From Extreme Position ( $x=-A$ ) back to Mean Position ( $x=0$ ): Distance = $A$

Calculate Total Distance: The total path length (distance) covered is the sum of these segments: $A + A + A + A = 4A$ . (Note: The net displacement is zero).

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