Energy of particle in simple harmonic motion

[Brian_D]

Homework Statement

The motion of a particle is described by $$x=(45 cm)[sin(pi*t+pi/6)]$$, with x in cm and t in seconds. At what time is the potential energy twice the kinetic energy?

Relevant Equations

$$potential energy=.5kx^2$$ $$kinetic energy=.5mv^2$$

Using the above equations for potential and kinetic energy, I figured that potential energy would be twice kinetic energy where $kx^2=2mv^2$. Solving this equation for v (where $k=mw^2$), we get $v=wx/sqrt 2$ Counting one complete cycle of the function on a graphing calculator, I found the period to be about t=57, so w is $2 pi/57# or about .11. So the potential energy would be twice the kinetic energy where$v=(.11/sqrt 2)x=.078x$To find the equation for velocity, I took the derivative of the motion equation$x=45[sin(pi*t+pi/6)]$. The derivative is:$45*pi*cos(pi*t + 1/6*Pi)$Finally, I set the velocity equation equal to the required expression for velocity obtained above, creating the equation:$45*Pi*cos(Pi*t + 1/6*Pi) = 0.078*x$. Solving this equation for t (using Maple), I get$t = (arccos(0.0005517371359*x) - Pi/6)/Pi$This is a different answer than the one given in the book answer key, which is$t=(0.14+n)s, t=(0.53+n)s, n an integer##. Your thoughts?

 

[Brian_D]

Sorry that the formatting of this got screwed up. I don't know how to recover my original text, so I hope you can follow what I said above.

 

[haruspex]

My first thought is to make your post legible by fixing the LaTeX.
You failed to double one of the #, put text inside that would be better outside, and omitted \ in front of sqrt, trig functions and pi (not Pi, please).
Please check the LaTeX before posting using the magnifying glass toggle at top right of the text entry area.

Homework Statement: The motion of a particle is described by $$x=(45 cm)[\sin(\pi*t+\pi/6)]$$, with x in cm and t in seconds. At what time is the potential energy twice the kinetic energy?

Relevant Equations: potential energy=$.5kx^2$ kinetic energy=$.5mv^2$

Using the above equations for potential and kinetic energy, I figured that potential energy would be twice kinetic energy where $kx^2=2mv^2$. Solving this equation for v (where $k=mw^2$), we get $v=wx/\sqrt 2$ Counting one complete cycle of the function on a graphing calculator, I found the period to be about t=57, so w is $2 \pi/57$ or about .11. So the potential energy would be twice the kinetic energy where $v=(.11/\sqrt 2)x=.078x$

To find the equation for velocity, I took the derivative of the motion equation $x=45[\sin(\pi*t+\pi/6)]$. The derivative is: $45*\pi*\cos(\pi*t + 1/6*\pi)$ Finally, I set the velocity equation equal to the required expression for velocity obtained above, creating the equation: $45*\pi*\cos(\pi*t + 1/6*\pi) = 0.078*x$. Solving this equation for t (using Maple), I get $t = (\arccos(0.0005517371359*x) - \pi/6)/\pi$

This is a different answer than the one given in the book answer key, which is $t=(0.14+n)s, t=(0.53+n)s$ , n an integer. Your thoughts?

 

[haruspex]

My second thought is that your work would be much clearer if you were to work purely algebraically and avoid plugging in numbers until the end. You should always work that way, it has many advantages.

 

[Brian_D]

Thank you for both the technical advice and your suggestion about not plugging numbers in until the end. I tried the latter in Maple and in this case did not even need to find a value for the angular frequency (w) in order for Maple to produce an answer, which was the same answer I got above. All of this will be helpful going forward. For the present problem, I would appreciate knowing where I'm going wrong.

 

[robphy]

The motion of a particle is described by $x=(45cm)[sin⁡(π∗t+π/6)]$, with x in cm and t in seconds. At what time is the potential energy twice the kinetic energy?

I found the period to be about t=57, so w is $2\pi/57$

Since $x=A\sin(\omega t + \phi)$, then $A=45{\rm\ cm}$, $\phi=\pi/6$, and $\omega=\pi$ (so, $\omega\neq 2\pi/57)$.

I figured that potential energy would be twice kinetic energy where $kx^2=2mv^2$

Write $U=\frac{1}{2}kx^2$ and $K=\frac{1}{2} mv^2$.
Since $\frac{1}{2}kA^2=U+K$, when $U=2K$, then $\frac{1}{2}kA^2=U+(\frac{1}{2}U)=\frac{3}{2}(\frac{1}{2}kx^2)$.
You can solve for the values of $x$.

 

[Orodruin]

The easiest way to solve this is to note that both energies oscillate between zero and the max amplitude with a sine or cosine squared with the same argument as the sine in x. Then the ratio of potential and kinetic energy is simply $\tan^2(\pi t + \pi/6) = 2$. Solve for $t$.

Your problem is you solved for $t$ without taking into account that $x$ also depends on $t$.

 

[robphy]

With my approach, you can solve for $x$, then solve for $t$
(or skip the solution for $x$ and go directly to $t$) in $\frac{2}{3}=\sin^2(\pi t +\frac{\pi}{6})$.

 

[SammyS]

Sorry that the formatting of this got screwed up. I don't know how to recover my original text, so I hope you can follow what I said above.

To edit a post you made, including the OP, click on the "Edit" feature to the lower left under your post.

 

[Orodruin]

To edit a post you made, including the OP, click on the "Edit" feature to the lower left under your post.

This is not available to everyone IIRC. Or at least I think different user categories have different time frames in which posts may be edited.

 

[berkeman]

I think the edit timeout for standard users (not Mentors or SAs) is 24 hours, but I'm not sure. In any case, if the OP wanted a post edited and it was past the timeout, they can always click "Report" on their post and ask the Mentors to help with the edits. In this case, it looks like @haruspex has fixed it up in his first reply, so no need to correct the OP.

 

[Brian_D]

Thank you, haruspex, robphy, Orodruin, SammyS and berkeman. Haruspex had suggested that the answer should include the tangent function, but I still didn't think of problem in terms of a ratio of the two energies until Orodruin explained it. I found robphy's approach to finding x helpful.

 

[haruspex]

Haruspex had suggested that the answer should include the tangent function

I later deleted that because I realised that because $\tan^2=\frac 1{\cos^2}-1$ it could be written either way. It just happened that the route I had found arrived at the tan form.

 

[Orodruin]

I later deleted that because I realised that because $\tan^2=\frac 1{\cos^2}-1$ it could be written either way. It just happened that the route I had found arrived at the tan form.

As is true for many trigonometric identities, one trig function can often be replaced by a function of another ...

It remains true for hyperbolic identities, my favourite one being the identity corresponding to $\tan^2=\frac 1{\cos^2}-1$, namely
$$
1 - \tanh^2(\theta) = \frac{1}{\cosh^2(\theta)}
$$
from which we can solve for $\cosh(\theta)$ to be
$$
\cosh(\theta) = \frac{1}{\sqrt{1 - \tanh^2(\theta)}}
$$
(now let $\tanh(\theta) = v/c$ and $\cosh(\theta) = \gamma$ ...)