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(edited 1 month ago)
https://isaacphysics.org/questions/graph_interpreting_6_beats?board=55c55eb6-d53d-452f-8071-b65868d206db&stage=a_level
Working is attached below, I’m stuck on parts an and b and would really appreciate some help. I’m sure I’m making some wrong assumptions, I just don’t know what they are. Thank you in advance.

part A requires you looking at the graph.

Youre sort of right although overly complex, but look at the time periods of the two sinusoids which are multiplied together (sum to product), so the solid and dashed lines, then f=1/T. Once you have those fm and Df, you can go back to f1 and f2 using the
f1/2 + f2/2 = fm
f1/2 - f2/2 = Df
which is easy to solve.

Note the =0s are wrong? Youve an expression and plotting the function. Youre not solving an equation.
(edited 1 month ago)
Original post by mqb2766
Youre sort of right although overly complex, but look at the time periods of the two sinusoids which are multiplied together (sum to product), so the solid and dashed lines, then f=1/T. Once you have those fm and Df, you can go back to f1 and f2 using the
f1+f2 = 2fm
f1-f2 = 2Df
which is easy to solve.
Note the =0s are wrong? Youve an expression and plotting the function. Youre not solving an equation.

Sorry to jump in but to clarify, are you saying to use the sun to angle formulae and equate that to 1/T in order to equate it to the freq, ie 1/0.05?
Original post by Hihobyu
Sorry to jump in but to clarify, are you saying to use the sun to angle formulae and equate that to 1/T in order to equate it to the freq, ie 1/0.05?

Sorry I mean 1/0.2, with 0.2 being the Time Period
Original post by Hihobyu
Here is my working. But I’m confused how to derive an equation for f_m, as the entire graph has the shape of the cos function, not the sin

The Df calc (large time period, small frequency is correct for the dashed envelope, though you may be overthinking the amount of equations/maths in the question. For fm calc (large frequency, small time period, solid curve), look at the sin() wave which lies inside (multiplied by) the dashed envelope formed by cos(). Time period is given by the axis crossing points ...

Once you have the sum to product identity to get
2Acos(2pi (f1-f2)/2 t)sin(2pi (f1+f2)/2 t)
You just use the f=1/T and read the T(s) off the graph and "solve" for f1 and f2.
(edited 1 month ago)
Original post by Hihobyu
These values are apparently wrong though. Could you show me where I went wrong?

wbf, its less about writing equations down and more about interpreting the graph. By looking at the axis crossing times, it should be easy to see that the time period of cos (dashed curve) is T=0.2, so Df=1/T=5. Just do a similar interpretation of the solid line (sin) where you note that even though its being "squashed" by cos, the axis crossing times are unaffected, so T=.... and fm=..... Then f1=..., f2=....

Dont write down any maths.
(edited 1 month ago)
What does wbf mean out of interest? I’ve seen you use it a couple of times

Without being funny. Its nbd (no big deal) but sometimes (often) the best way to solve maths problems is not to start writing down equations in the hope that algebra will save you, without having an idea of how things will work out.
Original post by mqb2766
wbf, its less about writing equations down and more about interpreting the graph. By looking at the axis crossing times, it should be easy to see that the time period of cos (dashed curve) is T=0.2, so Df=1/T=5. Just do a similar interpretation of the solid line (sin) where you note that even though its being "squashed" by cos, the axis crossing times are unaffected, so T=.... and fm=..... Then f1=..., f2=....
Dont write down any maths.

Thanks so much
Original post by mqb2766
wbf, its less about writing equations down and more about interpreting the graph. By looking at the axis crossing times, it should be easy to see that the time period of cos (dashed curve) is T=0.2, so Df=1/T=5. Just do a similar interpretation of the solid line (sin) where you note that even though its being "squashed" by cos, the axis crossing times are unaffected, so T=.... and fm=..... Then f1=..., f2=....
Dont write down any maths.

I don't really understand how can you get fm knowing T, T=0.2s, but how are you supposed to get fm? And how is it going to help you, does it have anything to do with fm=(f_1+f_2)/2 so if you know fm and you know f_1, you therefore know f_2.? Essentially my question is: how do you get fm?
Original post by mqb2766
Without being funny. Its nbd (no big deal) but sometimes (often) the best way to solve maths problems is not to start writing down equations in the hope that algebra will save you, without having an idea of how things will work out.

Also, by looking at the graph, shouldn't f_2 be the cos wave i.e. the dashed one. Hence shouldn't f_2 be 5Hz, as it makes one oscillation in 0.2s which is the time period?
Original post by Javier García
I don't really understand how can you get fm knowing T, T=0.2s, but how are you supposed to get fm? And how is it going to help you, does it have anything to do with fm=(f_1+f_2)/2 so if you know fm and you know f_1, you therefore know f_2.? Essentially my question is: how do you get fm?

cos is effective the dashed envelope and sin is effectively the (squashed) solid curve from which you can read off the time period and hence fm=1/T

The whole point of the question is that its hard to read off the time periods/frequencies when you have
sin() + sin()
however, if you use the sum to product identity and transform it into
cos()sin()
it should be easy as the cos is the envelope and sin is the fast oscillations within the envelope (hint 4).
(edited 1 month ago)
Original post by mqb2766
cos is effective the dashed envelope and sin is effectively the (squashed) solid curve from which you can read off the time period and hence fm=1/T
The whole point of the question is that its hard to read off the time periods/frequencies when you have
sin() + sin()
however, if you use the sum to product identity and transform it into
cos()sin()
it should be easy as the cos is the envelope and sin is the fast oscillations within the envelope (hint 4).
Thanks!!