Question

The motion of a particle is described by $x = 30\sin \left( {\pi t + \pi /6} \right)$, where $x$ is in $cm$ and $t$ in${\text{seconds}}$. The potential energy of the particle is twice the kinetic energy for the first time after $t = 0$ when the particle is at position _______ after _______ time.

Answer 1

Compare the equation of $x$ given in the question with the general equation of a wave, i.e., $x = A\sin (\omega t + \phi )$ to get the value of unknown variables such as $A$, $\omega$, $\phi$. Use the formulas for the potential energy and kinetic energy in the equality given in the question between these energies and solve the equation to get the value of $x$ (distance). Put this value of $x$ in the given equation to find the value of $t$ (time).

Formula Used:
Potential Energy = $\dfrac{1}{2}m{\omega ^2}{x^2}$
Kinetic Energy = $\dfrac{1}{2}m{\omega ^2}\left( {{A^2} - {x^2}} \right)$

Complete step by step answer:
As discussed in hint, compare the equation of $x$ provided in question that is $x = 30\sin \left( {\pi t + \pi /6} \right)$ with the general wave equation, $x = A\sin (\omega t + \phi )$ to get the values of $A$, $\omega$, $\phi$ beforehand.
We get, $A=30$, $\omega$ = $\pi$, $\phi$ = $\dfrac{\pi }{6}$.
Now, we are given that, ${\text{Potential Energy = 2}} \times {\text{Kinetic Energy}}$
After inputting the respective formulas, we get $\dfrac{1}{2}m{\omega ^2}{x^2}$ = 2 × $\dfrac{1}{2}m{\omega ^2}\left( {{A^2} - {x^2}} \right)$
On further solving, ${x^2} = 2 \times \left( {{A^2} - {x^2}} \right)$ ($\dfrac{1}{2}m{\omega ^2}$ gets cancelled)
Opening the brackets, ${x^2} = 2{A^2} - 2{x^2}$ (2 gets multiplied by both the variables)
$\Rightarrow 2{x^2} + {x^2} = 2{A^2}$ or, $3{x^2} = 2{A^2}$
Which gives, ${x^2} = \dfrac{2}{3}{A^2}$
Therefore $x = \sqrt {\dfrac{2}{3}{A^2}}$or, $x = \sqrt {\dfrac{2}{3}} A$.
$\Rightarrow x = \sqrt {\dfrac{2}{3}} \times 30$
Now, compare this value of $x$ with the given equation of $x$.
We get, $\sqrt {\dfrac{2}{3}} \times 30$ = $30\sin \left( {\pi t + \dfrac{\pi }{6}} \right)$
Further simplifying, $\sqrt {\dfrac{2}{3}} = \sin \left( {\pi t + \dfrac{\pi }{6}} \right)$ (30 gets cancelled)
$\Rightarrow {\sin ^{ - 1}}\sqrt {\dfrac{2}{3}} = \pi t + \dfrac{\pi }{6}$ (multiplying the equation by ${\sin ^{ - 1}}$).
$\Rightarrow s{\text{i}}{n^{ - 1}}\left( {\sqrt {\dfrac{2}{3}} } \right) - \dfrac{\pi }{6} = \pi t$
$\Rightarrow \dfrac{1}{\pi }\left[ {{{\sin }^{ - 1}}\left( {\sqrt {\dfrac{2}{3}} } \right) - \dfrac{\pi }{6}} \right] = t$
Therefore, $t = \dfrac{1}{\pi }{\sin ^{ - 1}}\left( {\sqrt {\dfrac{2}{3}} } \right) - \dfrac{1}{6}\sec$
On Further solving $x$, we get $x = 10\sqrt 6 cm$
So, the answer is: $10\sqrt 6 cm,\dfrac{1}{\pi }{\sin ^{ - 1}}\sqrt {\dfrac{2}{3}} - \dfrac{1}{6}\sec$

Note: Do not forget to put the units in the final answer. When you see that a formula is only making this equation more complex, drop the calculation and think of other formulas for the same. For example, in the above question, K.E. = $\dfrac{1}{2}m{v^2}$ will be of no use and hence should not be used.