Question 215: Oscillations

Question

A particle is executing linear S.H.M. starting from mean position. The ratio of the kinetic energy to the potential energy of the particle at a point of half the amplitude is

Accepted Answer

K.E. at half the amplitude position $\left(x=\frac{A}{2}ight)$
$$ =\frac{1}{2} m \omega^2\left(A^2-\frac{A^2}{4}ight)=\frac{3}{8} m \omega^2 A^2 $$
P.E. at $x=\frac{A}{2}=\frac{1}{2} m \omega^2\left(\frac{A}{2}ight)^2=\frac{1}{8} m \omega^2 A^2$
$	herefore \quad$ The required ratio
$$ \frac{	ext { K.E. }}{	ext { P.E. }}=\frac{\left(\frac{3}{8} m \omega^2 A^2ight)}{\left(\frac{1}{8} m \omega^2 A^2ight)}=\frac{3}{1} $$

Answer

$2: 1$

Answer

$3: 1$

Answer

$4: 1$

Answer

$8: 1$

Oscillations

A particle is executing linear S.H.M. starting from mean position. The ratio of the kinetic energy to the potential energy of the particle at a point of half the amplitude is

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