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,
5.25
(2) 2.5
(3) 4.5
(4) 7.5
(1)​

Answer 1

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

• $\rm{The\: amplitude\:af\:\:a\: pendulum=10cm}$
• $\rm{The\: pendulum\:is\:at\:a\: displacement=4cm}$

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

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

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

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

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

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

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

Here,

• P.E=Potential energy
• K.E=Kinetic energy
• A=Amplitude of pendulum
• X=Displacement (from the mean position)

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

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

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

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

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

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

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

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

$\rm\large{Hence,}$

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