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December 09, 2019, 11:21:49 am

Author Topic: Simple Harmonic Motion shifted equilibrium position.  (Read 294 times)  Share 

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Jefferson

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Simple Harmonic Motion shifted equilibrium position.
« on: April 28, 2019, 10:22:13 pm »
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In BOSTES 2015 Q13 (c) (not actually asking about the question itself)

They introduced the equation

v2 = n2 ( a2 - (x - c)2 )

Where 'a' is said to be the amplitude.



When I tried proving this result from
d/dx (v2/2) = - n2 ( x - c ), I am instead getting

v2  = n2 ( ( a + c )2 - x2 )


Could you please show me the formal proof to get to the equation
v2 = n2 ( a2 - ( x - c )2 ), where a is the amplitude.

Thank you!
« Last Edit: April 28, 2019, 10:26:02 pm by Jefferson »

RuiAce

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Re: Simple Harmonic Motion shifted equilibrium position.
« Reply #1 on: April 28, 2019, 10:35:15 pm »
+1
\[ \text{One way to buzzkill this proof is to actually recall that}\\ \boxed{\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)}+C}. \]
\begin{align*} \frac{d}{dx} \left( \frac{v^2}{2}\right) &= -n^2 (x-c)\\ \frac{v^2}{2} &= \frac{-n^2(x-c)^2}{2} + \frac{C}{2}\\ v^2 &= -n^2(x-c)^2+C \end{align*}
\[ \text{Recall that in SHM, }v=0\text{ when the particle is at its endpoints.}\\ \text{For a particle with centre of motion }c\text{ and amplitude }a,\\ \text{the endpoints of motion will be }x=c-a\text{ and }x=c+a.\]
\[ \text{Using either of these will then give}\\ 0 = -n^2a^2+C \implies \boxed{C=n^2a^2}. \]
\[ \text{Hence }v^2= -n^2(x-c)^2 + n^2a^2 = -n^2\left(a^2-(x-c)^2 \right)\]
Note that the proof should still work if we ordinarily integrate and get \( \frac{v^2}{2} = -n^2 \left( \frac{x^2}{2} - cx \right) \). My hunch, however, is that some completing the square would be needed later if we go down this route.

Jefferson

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Re: Simple Harmonic Motion shifted equilibrium position.
« Reply #2 on: April 28, 2019, 10:52:30 pm »
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\[ \text{One way to buzzkill this proof is to actually recall that}\\ \boxed{\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)}+C}. \]
\begin{align*} \frac{d}{dx} \left( \frac{v^2}{2}\right) &= -n^2 (x-c)\\ \frac{v^2}{2} &= \frac{-n^2(x-c)^2}{2} + \frac{C}{2}\\ v^2 &= -n^2(x-c)^2+C \end{align*}
\[ \text{Recall that in SHM, }v=0\text{ when the particle is at its endpoints.}\\ \text{For a particle with centre of motion }c\text{ and amplitude }a,\\ \text{the endpoints of motion will be }x=c-a\text{ and }x=c+a.\]
\[ \text{Using either of these will then give}\\ 0 = -n^2a^2+C \implies \boxed{C=n^2a^2}. \]
\[ \text{Hence }v^2= -n^2(x-c)^2 + n^2a^2 = -n^2\left(a^2-(x-c)^2 \right)\]
Note that the proof should still work if we ordinarily integrate and get \( \frac{v^2}{2} = -n^2 \left( \frac{x^2}{2} - cx \right) \). My hunch, however, is that some completing the square would be needed later if we go down this route.

Thank you so much! It seems I've made a huge blunder.
That makes a lot of sense!