A particle executes SHM of amplitude 25 cm and time period 12s. 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? (i.e. separation between points is 25 cm)  [Report your answer in seconds]

Let the displacement of the particle in SHM be given by $x\left(t\right)=A\sin (\omega t+\varphi ) \dots .\left(i\right)$

Where $A=25cm$ and $\omega =\frac{2\pi }{T}=\frac{2\pi }{3}rad{s}^{-1}$

Let us supposed that at time t = 0, the particle is at extreme position B. Setting x = A and t = 0 in Eq.

(i) We have $A=A\sin \varphi$

Giving $\varphi =\pi /2$

Putting $\varphi =\pi /2$ in Eq. we get $x\left(t\right)=A\cos \omega t\left(ii\right)$

Now let us say that the particle reaches point C at $t=t_1$ and point D at $t=t_2.$ At C, the displacement $x\left(t_1\right)=+12.5cm$ and at D, it is $x\left(t_2\right)=-12.5cm.$ So from (ii) we have $+12.5=25\cos \omega t_1$  And  $-12.5=25\cos \omega t_2$ Or  $\cos \omega t_1=+0.5or\omega t_1=\pi /3$ And $\cos \omega t_2=-0.5or\omega t_2=\frac{2\pi }{3}$

Hence $\omega (t_2-t_1)=\frac{2\pi }{3}-\frac{\pi }{3}=\frac{\pi }{3}$