Question

The amplitude of a particle executing SHM about $O$ is $10cm$. Choose the correct statements

• A. When the kinetic energy is $0.64$ of its maximum kinetic energy. its displacement is $6cm$ from $O$.
• B. When the displacement is $5cm$ from $O$ its kinetic energy is $0.75$ of its maximum kinetic energy
• C. Its total energy at any point is equal to its maximum kinetic energy
• D. Its velocity is half the maximum velocity when its displacement is half the maximum displacement.

Answer 1

Answer: (A), (B)

The correct options are
A When the kinetic energy is $0.64$ of its maximum kinetic energy. its displacement is $6cm$ from $O$.
B When the displacement is $5cm$ from $O$ its kinetic energy is $0.75$ of its maximum kinetic energy

Given $A=10cm$

Kinetic energy of particle executing $SHM$ at a displacement of $x$ from mean position is
$K=K_{max}(1-\frac{x^2}{A^2})\Rightarrow 0.64=1-\frac{x^2}{A^2}\Rightarrow x=0.6A\Rightarrow x=0.6\times 10\Rightarrow x=6cm.$
($A$ is correct)

When $x=5cm$
$K=K_{max}(1-\frac{5^2}{{10}^2})K=0.75K_{max}$ ($B$ is correct)

In general, the total energy is
$E=K_{max}+U_{min}$
Only if $U_{min}=0,$ the $E=K_{max}$ ($C$ is incorrect)


Now, velocity of particle executing $SHM$ at a displacement of $x$ from mean position is
$V=V_{max}\sqrt{1-\frac{x^2}{A^2}};$
if $x=\frac{A}{2}$
$V=V_0\sqrt{1-\frac{1}{4}}V=\frac{\sqrt{3}{V}_0}{2}$
($D$ is incorrect)