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प्रश्न
A particle is executing simple harmonic motion with amplitude A. When the ratio of its kinetic energy to the potential energy is `1/4`, its displacement from its mean position is ______.
विकल्प
`2/sqrt5`A
`sqrt3/2`A
`3/4`A
`1/4`A
MCQ
रिक्त स्थान भरें
उत्तर
A particle is executing simple harmonic motion with amplitude A. When the ratio of its kinetic energy to the potential energy is `1/4`, its displacement from its mean position is `underlinebb(2/sqrt5"A")`.
Explanation:
As we know
kinetic energy = `1/2"m"omega^2("A"^2-"x"^2)`
potential energy = `1/2"m"omega^2"x"^2`
∴ `(1/2"m"omega^2("A"^2-"x"^2))/(1/2"m"omega^2"x"^2)` = `1/4`
⇒ `("A"^2-"x"^2)/"x"^2` = `1/4`
4A2 - 4x2 = x2
⇒ x2 = `4/5`A2
⇒ x =`2/sqrt5`A.
shaalaa.com
Linear Simple Harmonic Motion (S.H.M.)
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