Question

A particle executes SHM of amplitude 25 cm and time periods 3 s. what is the minimum time required for the particle to move between two points located at 12.5 cm on either side of the mean position?

• A. 0.25 s
• B. 0.5 s
• C. 0.75 s
• D. 1.0 s

Answer 1

The correct option is B 0.5 s

Let A and B be the two extreme positions of the particle with O as the mean position. displace ments to the right of O are taken as positive while those to the left of O are taken as negative


Let the displacement of the particle is SHM be given by
$x\left(t\right)=Asin(\omega t+\varphi )$
where A = 25 cm and $\omega =\frac{2\pi }{T}=\frac{2\pi }{3}rad{s}^{-1}$ (i)

Let us suppose that at time t = 0, particle is at extreme position B. Setting x = A at t =0 in Eq. (i) we have
$A=Asin\varphi$
Giving $\varphi =\frac{\pi }{2}$
Putting $\varphi =\frac{\pi }{2}$ in Eq. (i), we get
$x\left(t\right)=Acos\omega t$
Where A = 25 cm
Now let us say that particle reaches point C at $t=t_1$ and point D at $t=t_2$. At C, the displacement $x\left(t_1\right)$ = + 12.5 cm and at D, it is $x\left(t_2\right)$ = 12.5 cm (see Fig 9.21). So from (ii) we have
$+12.5=25cos\omega t_{1}and-12.5=25cos\omega t_{2}orcos\omega t_1=+0.5or\omega t_1=\pi /3andcos\omega t_2=-0.5or\omega t_2=\frac{2\pi }{3}Hence\omega (t_1-t_2)=\frac{2\pi }{3}-\frac{\pi }{3}=\frac{\pi }{3}\therefore t_2-t_1=\frac{\pi }{3\omega }=\frac{T}{6}(\because \omega =\frac{2\pi }{T})Or{(t_2-t_1)}_{min}=\frac{3}{6}=0.5s$
Notice that $cos\omega t_2=-0.5evenfar{t}_2=\frac{4\pi }{3}$

This value of $t_2$ does not correspond to the minimum time because this is the time at which the particle , moving to left, reaches A and then
returns to D.