Question

.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,
(1) 5.25
(2) 2.5
(3) 4.5
(4) 7.5​

Answer 1

Answer:

4) 7.5

This is answer

Hope it helps uhh

Explanation:

Answer 2

$\sf\large\underline{Given:}$

$\sf{The\: amplitude\:af\:\:a\: pendulum=10cm}$

$\sf{The\: pendulum\:is\:at\:a\: displacement=4cm}$

$\sf\large\underline{To\: Find:}$

$\sf{The\: ratio\:of\:KE\:and\:PE\:at\:that\: point}$

$\sf\large\underline{Solution:}$

[By applying formula of kinetic energy and potential energy than simplify the ratio]

$\sf\large\underline{Formula\:Used:}$

$\sf{\implies K.E=\dfrac{1}{2}\:m\omega^2(A^2-x^2)}$

$\sf{\implies P.E=\dfrac{1}{2}\:m\omega^2\:x^2}$

Here,

P.E = Potential energy

K.E = Kinetic energy

A = Amplitude of pendulum

X = Displacement (from the mean position)

$\sf{\implies Ratio=\dfrac{K.E}{P.E}}$

$\sf{\implies K.E:P.E}$

$\sf{\implies \dfrac{1}{2}\:m\omega^2(A^2-x^2):\dfrac{1}{2}\:m\omega^2\:x^2}$

$\sf{\implies Ratio\:_{k.e:p.e}=\dfrac{A^2-x^2}{x^2}}$

$\sf{\implies Ratio\:_{k.e:p.e}=\dfrac{10^2-4^2}{4^2}}$

$\sf{\implies Ratio\:_{k.e:p.e}=\dfrac{100-16}{16}}$

$\sf{\implies Ratio\:_{k.e:p.e}=\dfrac{84}{16}}$

$\sf{\implies Ratio\:_{k.e:p.e}=\dfrac{21}{4}}$

$\sf\large{Hence,}$

$\sf{\implies Ratio\:_{kinetic\: energy: potential\: energy}=21:4}$