Question

A particle executes SHM with amplitude of $20cm$ and time period of $12sec$. What is the minimum time required for it to move between two points $10cm$ on either side of the mean position?

• A. $1$ sec
• B. $2$ sec
• C. $3$ sec
• D. $4$ sec

Answer 1

The correct option is B $2$ sec

$\text{Given:-}$ Particle amplitude (A) = 20 cm

Time period (T) = 12 sec

$\text{To find:-}$ The minimum time required

$\text{Solution:-}$

Time taken from complete 1 cycle = $t_0+t_0=2t_0$

$T=2t_0$

$x=Asin\omega t_0$

$10=Asin\frac{2\pi }{12}{t}_0$

as $T=\frac{2\pi }{\omega }$ So,

$\omega =\frac{2\pi }{12}$

Now,

$10=20sin\frac{2\pi }{12}{t}_0$

$\frac{10}{20}=sin\left(\frac{\pi }{6}\right)t_0$

$\frac{1}{2}=sin\left(\frac{\pi }{6}\right)t_0$

$sin\left(\frac{\pi }{6}\right)=sin\left(\frac{\pi }{6}\right)t_0$

as $\frac{1}{2}=sin\frac{\pi }{6}$

$\Rightarrow t_0=1$

T = $2t_0$

T = $2\times 1$ = 2sec.

$\text{Hence the correct option is B}$