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Simple harmonic motion

Screenshot_2016-04-12-14-36-52.png

I'm struggling with an mcq question here. Could anyone tell me why the maximum velocity is pi m/s? Urgent please. Thanks
Original post by Ridah
Screenshot_2016-04-12-14-36-52.png

I'm struggling with an mcq question here. Could anyone tell me why the maximum velocity is pi m/s? Urgent please. Thanks


They've swapped the labelling for the axes, BTW, but ... the magnitude of the max velocity for SHM is:

vmax=ωA=2πfA=2πATv_{\text{max}} = \omega A = 2\pi f A = \frac{2 \pi A}{T}

so just read the appropriate numbers off the graph, and plug them in.
Reply 2
Avatar for Zacken
Zacken
22
Original post by Ridah


I'm struggling with an mcq question here. Could anyone tell me why the maximum velocity is pi m/s? Urgent please. Thanks


From the graph x=π2cos⁡2t⇒x′=v=−πsin⁡2tx = \frac{\pi}{2}\cos 2t \Rightarrow x' = v = -\pi \sin 2t which is maximum precisely when sin⁡2t=−1\sin 2t = -1 and v=π ms−1v = \pi \, \mathrm{ms}^{-1}.
Original post by Zacken
From the graph x=π2cos⁡2t⇒x′=v=−πsin⁡2tx = \frac{\pi}{2}\cos 2t \Rightarrow x' = v = -\pi \sin 2t which is maximum precisely when sin⁡2t=−1\sin 2t = -1 and v=π ms−1v = \pi \, \mathrm{ms}^{-1}.


Where are you getting these numbers from? They don't look right to me.
Reply 4
Avatar for Zacken
Zacken
22
Original post by atsruser
Where are you getting these numbers from? They don't look right to me.


What's your AA and TT? I mostly hazarded a guess from the axes, the amplitude looked like π2\frac{\pi}{2} and the period like π\pi - I didn't see your answer before posting mine, whoops.
Reply 5
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Alexion
17
Original post by Zacken
What's your AA and TT? I mostly hazarded a guess from the axes, the amplitude looked like π2\frac{\pi}{2} and the period like π\pi - I didn't see your answer before posting mine, whoops.


The 'period''s definitely 3m. I'd guess the 'amplitude' is 1.5s
Reply 6
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Zacken
22
Original post by Alexion
The 'period''s definitely 3m. I'd guess the 'amplitude' is 1.5s


Huh? The axes should be swapped, no?
Reply 7
Avatar for Alexion
Alexion
17
Original post by Zacken
Huh? The axes should be swapped, no?


Yeah it's real confusing without the rest of the question to make it clearer :/
Original post by Alexion
Yeah it's real confusing without the rest of the question to make it clearer :/


Here is the question paper from what i can see:
Question 1:
Both of the responses here are correct, although i prefer the 2πfA2\pi fA method (because it works all the time every time)
unit 5 physics oscillations questions.pdf416.8KB
Original post by Zacken
Huh? The axes should be swapped, no?


You don't seem to have done that, as far as I can tell from your numbers. It seems to work out fine if you swap the axes then read off the values. As it stands, the graph doesn't make much sense - time only seems to exist between -2 and +2.
Reply 10
Avatar for Zacken
Zacken
22
Original post by atsruser
You don't seem to have done that, as far as I can tell from your numbers. It seems to work out fine if you swap the axes then read off the values. As it stands, the graph doesn't make much sense - time only seems to exist between -2 and +2.


I assumed it was a typo on the question setters part and mentally swapped the labels on the axes.
Original post by Zacken
I assumed it was a typo on the question setters part and mentally swapped the labels on the axes.


Err, yeah, but your numbers don't seem right though. T= 3 s, and A= 1.5 m, no?
Reply 12
Avatar for Zacken
Zacken
22
Original post by atsruser
Err, yeah, but your numbers don't seem right though. T= 3 s, and A= 1.5 m, no?


Right... I'll need to get myself declared legally blind.
Reply 13
Avatar for Ridah
Ridah
OP
Thank you so much.

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