GR 9277 Q no.17

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Joined: Thu Sep 03, 2009 8:26 am

GR 9277 Q no.17

Postby bullsye » Thu Sep 03, 2009 8:37 am

The lissajous figure for 2 SHM's whose frequencies are in the ratio 1:2 can be either a parabola(phase=90) or a figure of eight(phase=0,180). So, how do u answer this question in the traditional way when the phase difference is not given? :?:

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Joined: Sat Oct 20, 2012 1:56 pm

Re: GR 9277 Q no.17

Postby treadstone87 » Sat Oct 20, 2012 2:00 pm

Shouldn't the answer be (E)? It is given as (A) but if you take

omega_x = 2*omega
omega_y = omega

then the answer is (E).

The question states: Oscillator Y is connected to the vertical plates, while Oscillator X is connected to the horizontal plates.

To me, this suggests that Oscillator Y will control the x-signal and Oscillator X will control the y-signal. If one assumes this, then the plot should be a figure of eight.

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Joined: Tue Sep 01, 2015 10:37 am

Re: GR 9277 Q no.17

Postby Chiron » Tue Sep 01, 2015 12:38 pm

I realize that this is an older post, but I'll respond in the hopes it helps others studying for the exam.

In my opinion this question is very difficult, especially to solve during an exam. However, I think that one of the easiest ways to approach this may be to realize that options d and e are only possible if there is a phase difference between the waves. Thus, it cannot be d or e.

To separate between a, b, and c I think it's first helpful to write out the two equations:
x = A \sin(\omega t) where A is an arbitrary amplitude
y = B \sin(2 \omega t)

Thus, \dfrac{y}{x} = \dfrac{B}{A} \dfrac{\sin(2 \omega t)}{\sin(\omega t)} meaning that the slope is given by \dfrac{dy}{xx} = \dfrac{2 B \omega}{A \omega} \dfrac{\cos(2 \omega t)}{\cos(\omega t)}.
The importance here is that this slope is certainly not what would be expected for a normal sine or cosine curve. Thus, options b and c are excluded, leaving only a.

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