Physics, 01.07.2019 00:00 SoccerHalo

At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinetic and what fraction potential when the displacement is one third the amplitude

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Answer from: petriajack8375

The fraction of kinetic energy and fraction of Kinetic energy and Potential energy is mathematically given as

a) $x=\frac{A}{\sqrt{2}}$

b) F=8/9

c) Fp=1/9

Kinetic energy and Potential energy

Generally the equation for the Kinetic energy and Potential energy as is mathematically given as

KE = 1/2mw^2(A^2 - x^2)

and

PE = 1/2mw^2x^2

With

K.E and P.E equal we have $x=\frac{A}{\sqrt{2}}$

b) for when x = A/3

Generally the equation for the Kinetic energy is mathematically given as

KE = 1/2mw^2(A^2 - x^2)

herefore, the fraction of Kinetic energy is

$f = \frac{KE}{TE}$

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

f = 8/9

Therefore, fraction of PE is

Fp=1-f

Fp=1-(8/9)

Fp=1/9

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Answer from: MacenParisi

As we know that KE and PE is same at a given position

so we will have as a function of position given as

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

also the PE is given as function of position as

$PE = \frac{1}{2}m\omega^2x^2$

now it is given that

KE = PE

now we will have

$\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2$

A^2 - x^2 = x^2

2x^2 = A^2

$x = \frac{A}{\sqrt2}$

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

now to find the fraction of kinetic energy

$f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}$

$f = \frac{A^2 - (\frac{A}{3})^2}{A^2}$

$f_k = \frac{8}{9}$

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

$f_{PE} = 1 - f_k$

$f_{PE} = 1 - (8/9) = 1/9$

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

Answer from: Quest

This is the answer of your question

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