The amplitude of a simple pendulum is 10 cm. | vedclass

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Question

(A) The kinetic energy $(KE)$ of a simple harmonic oscillator is given by $KE = \frac{1}{2} m \omega^2 (A^2 - x^2)$.The potential energy $(PE)$ is giv

Vedclass · Accepted Answer

$5.25$

Vedclass · Answer

(A) The kinetic energy $(KE)$ of a simple harmonic oscillator is given by $KE = \frac{1}{2} m \omega^2 (A^2 - x^2)$.The potential energy $(PE)$ is given by $PE = \frac{1}{2} m \omega^2 x^2$.Here,$A = 10 \ cm$ and $x = 4 \ cm$.The ratio of kinetic energy to potential energy is $\frac{KE}{PE} = \frac{\frac{1}{2} m \omega^2 (A^2 - x^2)}{\frac{1}{2} m \omega^2 x^2} = \frac{A^2 - x^2}{x^2}$.Substituting the values: $\frac{KE}{PE} = \frac{10^2 - 4^2}{4^2} = \frac{100 - 16}{16} = \frac{84}{16}$.Calculating the result: $\frac{84}{16} = 5.25$.

The amplitude of a simple pendulum is $10 \ cm$. When the pendulum is at a displacement of $4 \ cm$ from the mean position,the ratio of kinetic and potential energies at that point is

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