Question

The amplitude of a particle executing SHM about $O$ is $10\text{cm}$. Then:

• A. When the K.E. is $0.64$ of its max. K.E., its displacement is $6\text{cm}$ from $O$.
• B. When the displacement is $5\text{cm}$ from $O$, its K.E. is $0.75$ of its max.P.E.
• C. Its total energy at any point is equal to its maximum K.E.
• D. Its velocity is half the maximum velocity when its displacement is half the maximum displacement.

Answer 1

Answer: (A), (B), (C)

The correct options are
A When the K.E. is $0.64$ of its max. K.E., its displacement is $6\text{cm}$ from $O$.
B When the displacement is $5\text{cm}$ from $O$, its K.E. is $0.75$ of its max.P.E.
C Its total energy at any point is equal to its maximum K.E.

Assume ideal S.H.M.:-
So, $\text{K.E.+P.E.}={\text{K.E}}_{\text{max}}$
At any position $x$, from mean position velocity $=\omega \sqrt{A^2-x^2}$
(A) $x=6\text{cm},A=10\text{cm}$
$v_0=\omega \times 10$
$V_{x=6\text{cm}}=\omega \sqrt{100-36}\Rightarrow \omega \times 8$
$\frac{{\text{KE}}_0}{{\text{KE}}_6}=\frac{100}{64}\Rightarrow 0.64$
(B) $x=5\text{cm},v=\omega \sqrt{100-25}\Rightarrow \omega \sqrt{75}$
$\frac{{\text{KE}}_5}{{\text{KE}}_0}=\frac{{\omega }^2\times 75}{{\omega }^2\times 100}\Rightarrow 0.75$