First slide

Simple hormonic motion

Question

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

Moderate

A

$0.6 \sec$
B

0.4 sec
C

0.3 sec
D

0.2 sec

Solution

Time taken by particle to move from x = 0 (mean position) to x = 4 (extreme position) = $\frac{T}{4}$ = $\frac{1.2}{4}$ = 0.3 s
Let t be the time taken by the particle to move from x = 0 to x = 2 cm
$y = a \sin ωt \Rightarrow 2 = 4 \sin \frac{2 π}{T} t \Rightarrow \frac{1}{2} = \sin \frac{2 π}{1.2} t$
$\Rightarrow \frac{π}{6} = \frac{2 π}{1.2} t \Rightarrow t = 0.1 s . Hence time to move from x = 2 to x = 4 will be equal to 0.3 - 0.1 = 0.2 s$
Hence total time to move from x = 2 to x = 4 and back again = $2 \times 0.2 = 0.4 \sec$

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