Question

A body performs simple harmonic oscillations along the straight ABCDE with C as the midpoint of AE.Its kinetic energies at B and D are each one fourth of its maximum value.If AE=2R the distance between B and D is

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

Answer 1

The correct option is C $\sqrt{3}R$

Since kinetic energy at B and D are equal, which means both points are equidistant from mean position, let say this distance is $x$. So the distance between B and D is $2x$, which we need to calculate.
Maximum value of kinetic energy= total energy$=\frac{1}{2}k{A}^2$
Kinetic energy at B = Kinetic energy at D $=\frac{1}{4}\times \frac{1}{2}k{A}^2=\frac{1}{8}k{A}^2$
$\Rightarrow$Potential energy at B = Potential energy at D=Total energy-kinetic energy $=\frac{1}{2}k{A}^2-\frac{1}{8}k{A}^2=\frac{3}{8}k{A}^2$
$\Rightarrow \frac{1}{2}k{x}^2=\frac{3}{8}k{A}^2$
$\Rightarrow x=\frac{\sqrt{3}}{2}A$
Here amplitude is $R$ so $x=\frac{\sqrt{3}}{2}R$
$\Rightarrow 2x=\sqrt{3}R$