Question

If a particle is executing $SHM$ on a straight line. $A$ and $B$ are two points at which its velocity is zero. It passes through a certain point $P$ ($AP<PB$) at successive intervals of $0.5$ sec and $1.5$ sec with a speed of $3$ m/s. The maximum speed of the particle is

• A. $3m/s$
• B. $3\sqrt{2}m/s$
• C. $6m/s$
• D. $3\sqrt{3}m/s$

Answer 1

Answer: (B)

$V=\omega \sqrt{A^2-x^2}$
$\omega \times t=\theta$

so,
$\omega \times 0.5=\theta$
$\omega \times 1.5=(2\pi -\theta )$

on adding the above equations;

$\omega \times 2=2\pi$
$\omega =\pi$
$\therefore \theta =\frac{\pi }{4}$
$x=Acos\theta$
$\frac{3=\pi \sqrt{A^2-\frac{A^2}{2}}}{V=\pi A}$
$\therefore V_{max}=3\sqrt{2}m/sec$