Question

If a body is executing simple harmonic motion and its current displacements is `sqrt3/2` times the amplitude from its mean position, then the ratio between potential energy and kinetic energy is:

विकल्प

• 3 : 2

• 2 : 3

• `sqrt3 : 1`

• 3 : 1

Answer 1

3 : 1

Explanation:

For displacement y = `(sqrt3 "A")/2`

PE = `1/2 "m"ω^2"y"^2 ⇒ 1/2"m"ω^2 [(sqrt3 "A")/2]^2`

KE = `1/2 "m"ω^2 ("A"^2 - "y"^2)^2 ⇒ 1/2 "m"ω^2 ("A"^2 - [(sqrt3"A")/2]^2)`

`"PE"/"KE" = (1/2 "m"ω^2 xx (3"A"^2)/4)/(1/2 "m"ω^2 "A"^2/4) = 3/1`