Question

The amplitude of a particle executing SHM about $O$ is $10cm$. Then

• A. when the K.E. is $0.64$ of its maximum K.E., its displacement is $6cm$ from $O$.
• B. when the displacement is $5cm$ from $O$, its K.E. is $0.75$ of its maximum 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 maximum K.E., its displacement is $6cm$ from $O$.
B when the displacement is $5cm$ from $O$, its K.E. is $0.75$ of its maximum P.E.
C its total energy at any point is equal to its maximum K.E.

We know that in SHM, total mechanical energy is conserved:
$ME=KE+PE$
$ME={KE}_{max}$
$ME={PE}_{max}$

Now, when $KE=0.64{KE}_{max}$ and $A=10cm$
By conservation of mechanical energy:
$PE=0.36{KE}_{max}=0.36{PE}_{max}$

We know that in SHM:
$PE=\frac{k{x}^2}{2}$
${PE}_{max}=\frac{k{A}^2}{2}$
$\frac{PE}{{PE}_{max}}=\frac{x^2}{A^2}$

which gives $x=6cm$

Now when $x=5cm$
$\frac{PE}{{PE}_{max}}=\frac{x^2}{A^2}$

which gives $PE=0.25{PE}_{max}$

Hence, $KE=0.75{KE}_{max}=0.75{PE}_{max}$