A body executes simple harmonic motion with an amplitude A. At what displacement from the mean position is the potential energy of the body is one fourth of its total energy ?

Question

A body executes simple harmonic motion with an amplitude A. At what displacement from the mean position is the potential energy of the body is one fourth of its total energy ? (A) A / 4 (B) A/2 (C) 3A/4 (D) Some other fraction of A

Tardigrade Admin · Accepted Answer

P.E =(1/2) M ω2 x2=(1/4) E=(1/4)((1/2) M ω2 A2) where total energy E =(1/2) M ω2 A2 ∴ x=(A/2)

Q. A body executes simple harmonic motion with an amplitude $A$ . At what displacement from the mean position is the potential energy of the body is one fourth of its total energy ?

A / 4

A/2

3A/4

Some other fraction of A

$P . E = \frac{1}{2} M ω^{2} x^{2} = \frac{1}{4} E = \frac{1}{4} (\frac{1}{2} M ω^{2} A^{2})$
where total energy $E = \frac{1}{2} M ω^{2} A^{2}$
$∴ x = \frac{A}{2}$

Q. A body executes simple harmonic motion with an amplitude A. At what displacement from the mean position is the potential energy of the body is one fourth of its total energy ?

A / 4

A/2

3A/4

Some other fraction of A

P.E=21​Mω2x2=41​E=41​(21​Mω2A2) where total energy E=21​Mω2A2 ∴x=2A​

Q. A body executes simple harmonic motion with an amplitude $A$ . At what displacement from the mean position is the potential energy of the body is one fourth of its total energy ?

$P . E = \frac{1}{2} M ω^{2} x^{2} = \frac{1}{4} E = \frac{1}{4} (\frac{1}{2} M ω^{2} A^{2})$
where total energy $E = \frac{1}{2} M ω^{2} A^{2}$
$∴ x = \frac{A}{2}$