A particle performs S.H.M. with amplitude 25 cm and period 3 s. The minimum time required for it to move between two points 12.5 cm on either side of the mean position is

Question

A particle performs S.H.M. with amplitude 25 cm and period 3 s. The minimum time required for it to move between two points 12.5 cm on either side of the mean position is (A) 0.6 s (B) 0.5 s (C) 0.4 s (D) 0.2 s

Tardigrade Admin · Accepted Answer

sin θ=(12.5/25) θ=(π/6) So, total angle to move =(2 π/6)=(π/3) It takes 3 s to rotate 360°. So, to rotate 60 t=(3/360) × 60 t =0.5 s

Q. A particle performs $S$ . $H$ . $M$ . with amplitude $25 c m$ and period $3 s$ . The minimum time required for it to move between two points $12.5 c m$ on either side of the mean position is

$0.6 s$

$0.5 s$

$0.4 s$

$0.2 s$

$sin θ = \frac{12.5}{25}$
$θ = \frac{π}{6}$
So, total angle to move $= \frac{2 π}{6} = \frac{π}{3}$
It takes 3 s to rotate $36 0^{\circ}$ . So, to rotate 60
$t = \frac{3}{360} \times 60$
$t = 0.5 s$

Q. A particle performs S.H.M. with amplitude 25cm and period 3s. The minimum time required for it to move between two points 12.5cm on either side of the mean position is

0.6s

0.5s

0.4s

0.2s