Help with SHM pls (1 Viewer)

sharoooooo · Aug 6, 2015

How do i do this?? please
Answer is A, but why? how??

Thanks

obliviousninja · Aug 6, 2015

Use the formula lol?

InteGrand · Aug 6, 2015

sharoooooo said:
How do i do this?? please
Answer is A, but why? how??

View attachment 32278

Thanks

$Since the period is $6$, we have $\frac{2\pi}{n}=6\Rightarrow n = \frac{2\pi}{6}=\frac{\pi}{3}$, where $n$ is the coefficient of $t$ in the argument of the trig. function. So we eliminate (C) and (D).$

$Also, $x=\int v(t)\text{ d}t$. We can see that if we try option (B), then $x=\int 5\cos \left(\frac{\pi}{3}t \right)\text{ d}t=\frac{5}{\frac{\pi}{3}}\sin \left(\frac{\pi}{3}t \right) +c$, and the amplitude of this wave is \emph{not} the required 5 (as the amplitude is instead $\frac{5}{\frac{\pi}{3}}$). Hence we eliminate (B). So (A) must be the answer. (We can check that using (A) and integrating, we indeed get an amplitude of 5, as required.)$

sharoooooo · Aug 6, 2015

InteGrand said:
$Since the period is $6$, we have $\frac{2\pi}{n}=6\Rightarrow n = \frac{2\pi}{6}=\frac{\pi}{3}$, where $n$ is the coefficient of $t$ in the argument of the trig. function. So we eliminate (C) and (D).$

$Also, $x=\int v(t)\text{ d}t$. We can see that if we try option (B), then $x=\int 5\cos \left(\frac{\pi}{3}t \right)\text{ d}t=\frac{5}{\frac{\pi}{3}}\sin \left(\frac{\pi}{3}t \right) +c$, and the amplitude of this wave is \emph{not} the required 5 (as the amplitude is instead $\frac{5}{\frac{\pi}{3}}$). Hence we eliminate (B). So (A) must be the answer. (We can check that using (A) and integrating, we indeed get an amplitude of 5, as required.)$

Thank you so much!!!

Help with SHM pls (1 Viewer)

sharoooooo

Active Member

obliviousninja

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InteGrand

Well-Known Member

sharoooooo

Active Member

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