Question

A body performs SHM along the straight line ABCDE with C is the midpoint of AE. Its kinetic energies at B and D are each one-fourth of its maximum value. If AE = 2A, then the distance between B and D is:

• A. A
• B. A$\sqrt{2}$
• C. A$\sqrt{3}$
• D. A$\sqrt{5}$

Answer 1

Answer: (C)

A$\sqrt{3}$

$KEmat=\frac{1}{2}m{\omega }^2{A}^2$
$KE=\frac{1}{2}m{\omega }^2(A^2-x^2)$
$K{E}_B=K{E}_D=\frac{KEmax}{4}$
$\therefore (A^2-x^2)=\frac{x^2}{4}$
$x=A\sqrt{\frac{3}{2}}$
$CD=Asin\theta$
$A\frac{\sqrt{3}}{2}=Asin\theta$
$\theta =60$
$BD=2CD$
$=2\times \frac{\sqrt{3}A}{2}$
$=\sqrt{3}A$