Question

Assuming the particle to be stationary at the moment $t=0$, find the distance covered by the particle between two successive stops

• A. $\Delta s=\frac{2\overrightarrow{F}}{m{\omega }^2}$
• B. $\Delta s=\frac{8\overrightarrow{F}}{m{\omega }^2}$
• C. $\Delta s=\frac{4\overrightarrow{F}}{m{\omega }^2}$
• D. $\Delta s=\frac{3\overrightarrow{F}}{m{\omega }^2}$

Answer 1

The correct option is B $\Delta s=\frac{8\overrightarrow{F}}{m{\omega }^2}$

Let the force be
$\overrightarrow{F}=F\left[cos\right(\omega t)i+sin(\omega t\left)j\right]$
$\Rightarrow Accelertion\overrightarrow{a}=\frac{F}{m}\left[cos\right(\omega t)i+sin(\omega t\left)j\right]$
$\Rightarrow \frac{d\overrightarrow{v}}{dt}=\frac{F}{m}\left[cos\right(\omega t)i+sin(\omega t\left)j\right]$
$\Rightarrow {\int }_0^{\overrightarrow{v}}d\overrightarrow{v}=\frac{F}{m}{\int }_0^t\left[cos\right(\omega t)i+sin(\omega t\left)j\right]dt$
$\Rightarrow \overrightarrow{v}=\frac{F}{m\omega }\left[sin\right(\omega t)i+(1-cos\left(\omega t\right)j\left)\right]$
The particle was stationary at tme $t=0$. For particle to be stationary again, both components of velocity must be zero. $\Rightarrow sin\left(\omega t\right)=0and1-cos\left(\omega t\right)=0$
$\Rightarrow t=\frac{n\pi }{\omega }andt=\frac{2n^{\prime }\pi }{\omega }$
Taking intersection
$\Rightarrow t=\frac{2\pi }{\omega }$
Thus the particle wil again be stationary at $t=\frac{2\pi }{\omega }$
Speed, $v=\sqrt{v_x^2+v_y^2}$
$\Rightarrow v=\frac{F}{m\omega }\sqrt{2-2cos\left(\omega t\right)}=\frac{2F}{m\omega }sin\left(\frac{\omega t}{2}\right)$
Distance covered between successive stops, $\Delta s={\int }_0^{\frac{2\pi }{\omega }}vdt=\frac{2F}{m\omega }{\int }_0^{\frac{2\pi }{\omega }}sin\left(\frac{\omega t}{2}\right)dt$
$\Rightarrow \Delta s=\frac{8F}{m{\omega }^2}$
Thus correct option is (b)