Question

For what displacement the P.E becomes 1/4 of its maximum value?

Answer 1

For free falling body.

let v² = u²+2*g*h’

as u =0

Now according to the question

½*m*v²=¼mgh.
where h = initial height

½(2*gh')=¼gh

or h'=h/4

Answer 2

Correct Question is "For what displacement the P.E of SHM becomes 1/4 of its maximum value."

Answer:

The displacement is $x = \pm \frac{A}{2}$ in which potential energy becomes 1/4 of its maximum value.

Explanation:

For simple harmonic motion,total energy is $\frac{1}{2} m \omega^2 A^{2}$ [wher,A is the amplitude i.e. the maximum displacement, $\omega$ is the angular velocity of the body, m is mass of a body]

Now,potential energy of a body is $\frac{1}{2} m \omega^2 x^{2}$ [ $x$ is displacement]

According to the question,

    Potential Energy of SHM becomes 1/4 of its maximum value

       ∴  $\frac{1}{2} m \omega^2 x^{2}$ = $\frac{1}{4} \times \frac{1}{2} m \omega^2 A^{2}$

              → $x^{2} = \frac{1}{4} A^{2}$

              → $x = \pm \frac{A}{2}$

∴The displacement is $x = \pm \frac{A}{2}$ in which potential energy becomes 1/4 of its maximum value.

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