Question

$A$ and $B$ are two points on the path of a particle executing SHM in a straight line, at which its velocity is zero. Starting from a certain point $X[AX<BX]$, the particle crosses the point $X$ in successive intervals of $2$ seconds and $4$ seconds with a speed of $5\text{m/s}$, then

• A. The amplitude of oscillation is $\frac{10\sqrt{m}}{\pi }\text{m}$
• B. The maximum speed of the particle is $\frac{5}{\sqrt{3}}\text{m/s}$
• C. The maximum speed of the particle is $\frac{10}{\sqrt{3}}\text{m/s}$
• D. The amplitude of oscillation is $\frac{5\sqrt{3}}{\pi }\text{m}$

Answer 1

Answer: (A), (C)

The maximum speed of the particle is $\frac{10}{\sqrt{3}}\text{m/s}$

$T=2+4=6\text{s}$
$\Rightarrow \omega =\frac{2\pi }{T}=\frac{\pi }{3}$
$\because x=A\cos \omega t$
$\Rightarrow x=A\times \cos (\frac{\pi }{3}\times 1)\Rightarrow x=\frac{A}{2}$
$\therefore v=\omega \sqrt{A^2-x^2}=\frac{\pi }{3}\sqrt{\frac{3A^2}{4}}$
$\Rightarrow 5=\frac{\pi }{3}\times \frac{\sqrt{3}A}{2}$
$\Rightarrow A=\frac{10\sqrt{3}}{\pi }\text{m}$
$\therefore v_{\text{max}}=\omega A=\frac{\pi }{3}\times \frac{10\sqrt{3}}{\pi }=\frac{10}{\sqrt{3}}\text{m/s}$