A simple pendulum is oscillating with amplitude ‘A’ and angular frequency ‘ω’. At displacement ‘x’ from mean position, the ratio of kinetic energy to potential energy is

Question

A simple pendulum is oscillating with amplitude ‘A’ and angular frequency ‘ω’. At displacement ‘x’ from mean position, the ratio of kinetic energy to potential energy is (A) (x2/A2-x2) (B) (x2-A2/x2) (C) (A2-x2/x2) (D) (A-x/x)

Tardigrade Admin · Accepted Answer

Kinetic energy of pendulum oscillating with amplitude

A

and angular frequency

ω

at displacement

x

from mean position is

K E = \frac{1}{2} k (A^{2} - x^{2})

Potential energy of pendulum at displacement

x

from mean position is

PE = \frac{1}{2} k x^{2}

∴ \frac{K E}{PE} = \frac{\frac{1}{2} k ( A ^{2} - x ^{2} )}{\frac{1}{2} k x ^{2}} = \frac{A ^{2} - x ^{2}}{x ^{2}}

Q. A simple pendulum is oscillating with amplitude ‘ $A$ ’ and angular frequency ‘ $ω$ ’. At displacement ‘ $x$ ’ from mean position, the ratio of kinetic energy to potential energy is

$\frac{x ^{2}}{A ^{2} - x ^{2}}$

$\frac{x ^{2} - A ^{2}}{x ^{2}}$

$\frac{A ^{2} - x ^{2}}{x ^{2}}$

$\frac{A - x}{x}$

Q. A simple pendulum is oscillating with amplitude ‘A’ and angular frequency ‘ω’. At displacement ‘x’ from mean position, the ratio of kinetic energy to potential energy is

A2−x2x2​

x2x2−A2​

x2A2−x2​

xA−x​

Kinetic energy of pendulum oscillating with amplitude A and angular frequency ω at displacement x from mean position is KE=21​k(A2−x2) Potential energy of pendulum at displacement x from mean position is PE=21​kx2 ∴PEKE​=21​kx221​k(A2−x2)​=x2A2−x2​

Q. A simple pendulum is oscillating with amplitude ‘ $A$ ’ and angular frequency ‘ $ω$ ’. At displacement ‘ $x$ ’ from mean position, the ratio of kinetic energy to potential energy is

$\frac{x ^{2}}{A ^{2} - x ^{2}}$

$\frac{x ^{2} - A ^{2}}{x ^{2}}$

$\frac{A ^{2} - x ^{2}}{x ^{2}}$

$\frac{A - x}{x}$