Question

The motion of a particle is given by , where t is in $s$. What is the first time at which the kinetic energy is twice the potential energy?

Short answer

$95$is the timeduring which theK.E. is twice theP.E..

Step-by-step solution

Determine the kinetic energy.

We would prefer to know whenever the overall energy balances the mechanical energy.

$K=2U$

In basic periodic movement, , an object is in motion with such a velocity $V$, and an accelerator $a$ in periodic motion, during which parameters $a,v$, $x$are harmonic functions of time $t$. The frequency $a$and phase $\varphi$ of this oscillation couldbecalculated using an equation.

$x\left(t\right)=A\cos (\omega t+\varphi )$

When $a$indicates amplitude, $\omega$indicates frequency response, and $t$signifies time. In the problem $37$, we were provided the particle's motion equation by

$x\left(t\right)=\left(25\mathrm{cm}\right)\cos \left(10t\right)$

When we compare equation $\left(2\right)$to equation $\left(1\right)$, we discover that $A=25\mathrm{cm},\omega =10\mathrm{rad}/\mathrm{s}$

and $\varphi =0$Next, search out the equation $\left(2\right)$,$t$to find the speed.$v\left(t\right)=\frac{dx}{dt}$

$v\left(t\right)=\frac{dx}{dt}$

$=-\left(250\mathrm{cm}\right)\sin \left(10t\right)$

The mechanical energy is the mass multiplied either by quadratic velocity. $v\left(t\right),s$one could construct a formula for the kinetic energy $k$in respect time $t$by

$\begin{array}{r}K=\frac{1}{2}m{v}^2\end{array}$

$=\frac{1}{2}m[-(250\mathrm{cm})\sin (10t\left)\right]$

$=\left(31.25\times {10}^3{\mathrm{cm}}^2\right)m{\sin }^2\left(10t\right)$

Expression of potential.

The mechanical energy in terms of the time $t$could be expressed as next

$U=\frac{1}{2}k{x}^2$

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

$=\frac{1}{2}m(10\mathrm{rad}/\mathrm{s}{)}^2((25\mathrm{cm})\cos \left(10t\right){)}^2$

$=\left(31.25\times {10}^3{\mathrm{cm}}^2\right)m{\cos }^2\left(10t\right)$

As we mentioned above, we would like$t$when $K=2U$, so we are able to find the worth of the time $t$by

$k=2u$

$\left(31.25\times {10}^3{\mathrm{cm}}^2\right)m{\sin }^2\left(10t\right)=2\left(31.25\times {10}^3{\mathrm{cm}}^2\right)m{\cos }^2\left(10t\right)$

${\left(\frac{\sin 10t}{\cos 10t}\right)}^2=2$

${\tan }^2\left(10t\right)=2$

$\tan \left(10t\right)=\sqrt{2}$

$10t={\tan }^{-1}\sqrt{2}$

$t=\frac{{\tan }^{-1}\sqrt{2}}{10}$

$t=0.095\mathrm{s}$

In the final step, we converted the calculator mode into rad. $t=95ms$