A particle is performing simple harmonic motion along x-axis with amplitude 4 cm and time period 1.2 s. The minimum time taken by the-particle to move from x=+2cm to x=+4and back again is given by

A
0.6 sec
B
0.4 sec
C
0.3 sec
D
0.2 sec

Solution:

short cut : t = $\frac{2 T}{6} = \frac{T}{3}$ =0.4 sec

Time taken from O to A
$= \frac{T}{8} = \frac{1 \cdot 2}{8} = \frac{0 \cdot 3}{2} Time taken from O to B . = \frac{T}{4} = \frac{1 \cdot 2}{4} = 0 \cdot 3 Tlme taken from A to B = (0 \cdot 3 - \frac{0 \cdot 3}{2}) = \frac{0 \cdot 3}{2} Total time taken = 2 \times \frac{0 \cdot 3}{2} = 0 \cdot 3 \sec$

A particle is performing simple harmonic motion along x-axis with amplitude 4 cm and time period 1.2 s. The minimum time taken by the-particle to move from $x = + 2 cm to x = + 4$ and back again is given by

Solution:

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