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Q.

What is the ratio between the potential energy (U) and the total energy (E) of a particle, and kinetic energy (K) and total energy (E) of a particle, executing $S H M$ , when its displacement is half of its amplitude?

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a

$\frac{K}{E} = \frac{1}{2}$

b

$\frac{U}{E} = \frac{1}{4}$

c

$\frac{U}{E} = \frac{1}{2}$

d

$\frac{K}{E} = \frac{3}{4}$

answer is B, D.

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Detailed Solution

$E = \frac{1}{2} m ω^{2} A^{2} y = \frac{A}{2} U = \frac{1}{2} m ω^{2} y^{2} - \frac{1}{2} m ω^{2} {(\frac{A}{2})}^{2} = \frac{E}{4} y = \frac{A}{2} K = \frac{1}{2} m v^{2} = \frac{1}{2} m ω^{2} (A^{2} - y^{2}) = \frac{1}{2} m ω^{2} (A^{2} - \frac{A^{2}}{4}) = \frac{3 E}{4}$

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