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

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

Answer 1

The correct option is B

0.4 sec

Time taken by particle to move from x=0 (mean position) to

x = 4 (extreme position) = $\frac{T}{4}=\frac{1.2}{4}=0.3s$

Let t be the time taken by the particle to move from x = 0 to x = 2 cm

y = a sin $\omega$ t $\Rightarrow 2=4sin\frac{2\pi }{T}r\Rightarrow \frac{1}{2}=sin\frac{2\pi }{1.2}t$
$\Rightarrow \frac{\pi }{6}=\frac{2\pi }{1.2}t$ = 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 x 0.2 = 0.4 sec