Question

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

• A. When the K.E. is 0.64 of its max. K.E. its displacement is 6cm from O.
• B. When the displacement is 5 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 6cm from O.
B When the displacement is 5 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.

Amplitude$=10cm$

$KE=\frac{1}{2}m{\omega }^2(A^2-x^2)$

${KE}_{max}=\frac{1}{2}m{\omega }^2{A}^2$

When particle will be at mean position,

$0.64=\frac{KE}{{KE}_{max}}\Rightarrow 0.64=\frac{\frac{1}{2}m{\omega }^2(A^2-x^2)}{\frac{1}{2}m{\omega }^2{A}^2}$

$\Rightarrow 0.64=\frac{(A^2-x^2)}{A^2}$

$\Rightarrow x^2=100-(25\times 0.64\times 4)$

$x=\sqrt{36}cmx=6cm$

${PE}_{max}={KE}_{max}$

$0.75=\frac{A^2-x^2}{A^2}{x}^2=100-75x=5cm$ from mean position.

$KE+PE=TE{KE}_{max}+0=TE$

[PE becomes zero when KE is maximum]