Question

A particle executes SHM of amplitude $25\text{cm}$ and time period $3\text{s}$. What is the minimum time required for the particle to move between two points located at $12.5\text{cm}$ from the mean position on either side?

• A. $0.25\text{s}$
• B. $0.5\text{s}$
• C. $0.75\text{s}$
• D. $1.0\text{s}$

Answer 1

The correct option is B $0.5\text{s}$


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

Let us suppose that at time $t=0$, the particle is at extreme position $B$. Setting $x=A$ at $t=0$ in Eq. (i), we have
$A=A\sin \varphi$
giving $\varphi =\frac{\pi }{2}$
Putting $\varphi =\frac{\pi }{2}$ in Eq. (i), 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 $D$ immediately after at $t=t_2$. At $C$, the displacement
$x\left(t_1\right)=+12.5\text{cm}$ and at $D$, it is
$x\left(t_2\right)=-12.5\text{cm}$

So from (ii), we have
$+12.5=25\cos \omega t_1$
and $-12.5=25\cos \omega t_2$
or $\cos \omega t_1=+0.5$ or $\omega t_1=\frac{\pi }{3}$
and $\cos \omega t_2=-0.5$ or $\omega t_2=\frac{2\pi }{3}$

Hence, $\omega (t_2-t_1)=\frac{2\pi }{3}-\frac{\pi }{3}=\frac{\pi }{3}$
$\therefore t_2-t_1=\frac{\pi }{3\omega }=\frac{T}{6}$ $(\because \omega =\frac{2\pi }{T})$
i.e $(t_2-t_1{)}_{\text{min}}=\frac{3}{6}=0.5\text{s}$ $(\because T=3\text{s})$

Notice that $\cos \omega t_2=-0.5$ even for $t_2=\frac{4\pi }{3}$. This value of $t_2$ does not correspond to the minimum time because this is the time at which the particle, after reaching $A$ returns to $D$.