Question

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

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

Answer 1

The correct option is B

0.4 s

Let the displacement of the particle be given by $x=Asin\left(\frac{2\pi t}{T}\right)$ where A = 4 cm and T = 1.2 s.
If $t_1$ is the time taken by the particle to move from x = 0 to x = 2 cm, then
$2=4sin\left(\frac{2\pi t_1}{T}\right)$
$\Rightarrow t_1=\frac{T}{12}$
If $t_2$ is the time taken to move from x = 0 to x = 4 cm, then
$4=4sin\left(\frac{2\pi t_2}{T}\right)$
$\Rightarrow t_2=\frac{T}{4}$
Now, time taken to move from x = 2 cm to x = 4 cm is $t_2-t_1=\frac{T}{4}-\frac{T}{12}=\frac{T}{6}=\frac{1.2s}{6}=0.2s$. Therefore, time taken by the particle to move from x = 2 cm to x = 4 cm and back is $0.4s$ and the correct choice is (b).