Engineering Amplitude of superposition of forces

[rc2008]

Homework Statement

Find the amplitude of forces of superposition of 300$\cos13πt$ , 500$\sin6.4 πt$ + 300$\sin 13πt$ +600$\cos 18.8 πt$

Relevant Equations

Find the amplitude of forces of superposition of 300cos13πt, 500sin6.4 πt + 300sin 13πt +600cos 18.8 πt

My answer is $\sqrt((300^2)+300^2)$ = 424 after superimposing both $cos13πt$ and $sin13πt$, After which, the overall of amplitude of superimposed of all 4 forces are $\sqrt( (434^2)+ (500^2) + (600^2))$ = 893, correct me if I am wrong

 

[BvU]

I agree with the 424, but what is your reasoning to add the others in quadrature ?

$\$

 

[rc2008]

I agree with the 424, but what is your reasoning to add the others in quadrature ?

$\$

My reasoining is superimpose the Asin and Bcos curve having the same frequqncy first, then I would get resultant curve of $\sqrt(A^2+B^2)$ $\sin(θ+φ)$, then I only superimpose the resultant curve with other cos curve and sin cruve having different frequnecy, correct me if I am wrong.

 

[BvU]

My reasoining is superimpose the Asin and Bcos curve having the same frequqncy first, then I would get resultant curve of $\sqrt(A^2+B^2)$ $\sin(θ+φ)$, then I only superimpose the resultant curve with other cos curve and sin cruve having different frequnecy, correct me if I am wrong.

The plot in #2 shows function values of + and -1429, so you cannot be right ...

Same situation as here

What do you get for the amplitude of $\cos x+\sin x\$ ? For $\cos x+\sin (2x)\$ ?

For the first one, $\sqrt 2$ is right, but for the second one you didn't answer (and neither 2 nor $\sqrt 2$ is right!)

Make some plots yourself to find out how this works !

For lazy PFers: no work involved !

$\$

 

[Steve4Physics]

Homework Statement: Find the amplitude of forces of superposition of 300$\cos13πt$ , 500$\sin6.4 πt$ + 300$\sin 13πt$ +600$\cos 18.8 πt$

Presumably there is a simple typo' so that
300$\cos13πt$ , 500$\sin6.4 πt$ + 300$\sin 13πt$ +600$\cos 18.8 πt$
should be
300$\cos13πt$ + 500$\sin6.4 πt$ + 300$\sin 13πt$ + 600$\cos 18.8 πt$

To add to what @BvU has already said, note that at t=0 the cosine terms are all 1 and the sine terms are all 0, so the total force at t=0 is
300*1 + 500*0 + 300*0 + 600*1 = 900.
This exceeds your calculated amplitude!

Your method of amplitude-addition generally only works for 2 terms with the same frequency and a phase difference of $\frac {\pi} 2$.

It looks like the problem can only be solved numerically - if this is course-work you are probably expected to do it this way. What course are you on?

 

[rc2008]

The plot in #2 shows function values of + and -1429, so you cannot be right ...

Same situation as here

For the first one, $\sqrt 2$ is right, but for the second one you didn't answer (and neither 2 nor $\sqrt 2$ is right!)

Make some plots yourself to find out how this works !

For lazy PFers: no work involved !

$\$

Presumably there is a simple typo' so that
300$\cos13πt$ , 500$\sin6.4 πt$ + 300$\sin 13πt$ +600$\cos 18.8 πt$
should be
300$\cos13πt$ + 500$\sin6.4 πt$ + 300$\sin 13πt$ + 600$\cos 18.8 πt$

To add to what @BvU has already said, note that at t=0 the cosine terms are all 1 and the sine terms are all 0, so the total force at t=0 is
300*1 + 500*0 + 300*0 + 600*1 = 900.
This exceeds your calculated amplitude!

Your method of amplitude-addition generally only works for 2 terms with the same frequency and a phase difference of $\frac {\pi} 2$.

It looks like the problem can only be solved numerically - if this is course-work you are probably expected to do it this way. What course are you on?

Thanks, looks like I can only use pythagora's theorem for cos and sin curve with same frequency. for superposition of sin + sin / cos+ cos / sin+cos curve with different frequqncy, i can only just simply add up to get the magnitude of superimposed curve, correct me if I am wrong. 424+500+6001524, still exceed 1424 as @BvU pointed out.

 

[Steve4Physics]

Thanks, looks like I can only use pythagora's theorem for cos and sin curve with same frequency.

To use the $\sqrt {A^2+B^2}$ 'rule' the frequencies must be the same and also the phase difference must be an odd multiple of $\frac {\pi} 2$. For example it works for $3\sin(2t) + 4\cos(2t)$ or for $5\sin(11t + \frac {7\pi}2) + 2\sin(11t)$.

for superposition of sin + sin / cos+ cos / sin+cos curve with different frequqncy, i can only just simply add up to get the magnitude of superimposed curve, correct me if I am wrong. 424+500+6001524, still exceed 1424 as @BvU pointed out.

If you add up the individual amplitudes (424+500+600=1524) this gives the upper limit for the final amplitude - so you know the final amplitude cannot be greater than 1524. This could be a useful check once you have an answer but it does not help you to find the answer.

To find the amplitude you could use calculus (to find a maximum) but you will usually not end up with something that can be solved analytically. Finding the amplitude will usually require a numerical method (e.g. using MatLab). If only an approximate value is required, then it can be read-off a graph such as @BvU 's.

You never answered my question about what course you are taking!

 

[rc2008]

To use the $\sqrt {A^2+B^2}$ 'rule' the frequencies must be the same and also the phase difference must be an odd multiple of $\frac {\pi} 2$. For example it works for $3\sin(2t) + 4\cos(2t)$ or for $5\sin(11t + \frac {7\pi}2) + 2\sin(11t)$.


If you add up the individual amplitudes (424+500+600=1524) this gives the upper limit for the final amplitude - so you know the final amplitude cannot be greater than 1524. This could be a useful check once you have an answer but it does not help you to find the answer.

To find the amplitude you could use calculus (to find a maximum) but you will usually not end up with something that can be solved analytically. Finding the amplitude will usually require a numerical method (e.g. using MatLab). If only an approximate value is required, then it can be read-off a graph such as @BvU 's.

You never answered my question about what course you are taking!

Thanks for the info, this is related to dynamics courses and not mathematics courses. So , the approx value shall be good enought. I am struggling to find out the max amplitude of superposition of forces of different function and different frequency. For the $\sqrt {A^2+B^2}$ rule , does it work for $5\cos(11t + \frac {7\pi}2) + 2\sin(11t)$ ?

 

[FactChecker]

Finding the amplitude will usually require a numerical method (e.g. using MatLab). If only an approximate value is required, then it can be read-off a graph such as @BvU 's.

Good point. But he should also be warned that this function has many local maximums that are far below the global maximum. That makes it difficult to find the global maximum using numerical methods. I would suggest using some analysis of the function to determine if there is or is not a common frequency that would give a lower global maximum than 1524. Graphically, it looks like the global maximum is under 1500.

 

[rc2008]

Good point. But he should also be warned that this function has many local maximums that are far below the global maximum. That makes it difficult to find the global maximum using numerical methods. I would suggest using some analysis of the function to determine if there is or is not a common frequency that would give a lower global maximum than 1524. Graphically, it looks like the global maximum is under 1500.

Thanks, I do not have suffcient knowldege on mathematics. For this kind of superpostion of curve, is there any shortcut formula to find the amplitude of superimposed curve other than more comprehensive analysis like Matlab?

 

[Steve4Physics]

Thanks for the info, this is related to dynamics courses and not mathematics courses. So , the approx value shall be good enought. I am struggling to find out the max amplitude of superposition of forces of different function and different frequency. For the $\sqrt {A^2+B^2}$ rule , does it work for $5\cos(11t + \frac {7\pi}2) + 2\sin(11t)$ ?

No. The 2 'waves' to be added must in 'quadrature'. That means there must be a quarter-cycle shift ($\frac {\pi}2$ radians) between the 2 'waves' (or, more generally, a shift of an odd number of quarter cycles).

For example, $\sin x$ and $\cos x$ are in quadrature. But remember that $\sin x = \cos(x- \frac {\pi}2)$. So $\sin x$ is not in quadrature wirh $\cos(x- \frac {\pi}2)$.

You can use sites such as https://www.desmos.com/calculator to draw the graphs. Or use a graphing calculator or laptop+software or similar.

Sketch the graphs of $y=5\cos(11t + \frac {7\pi}2)$ and $y=2\sin(11t)$ for yourself. You will see the graphs are in phase, not in quadrature.

Generally, if the 'waves' do not have the same frequency and/or are not in quadrature, there is no shortcut formula/method I know of. Drawing the graph would be simplest/quickest method in this situation (IMO).

Edit: An advantage of drawing the graph is that you will see the local maxima (as well as the global maxima) as noted by @FactChecker. This could be very useful information, especially if some local maxima are large.

 

[rc2008]

No. The 2 'waves' to be added must in 'quadrature'. That means there must be a quarter-cycle shift ($\frac {\pi}2$ radians) between the 2 'waves' (or, more generally, a shift of an odd number of quarter cycles).

For example, $\sin x$ and $\cos x$ are in quadrature. But remember that $\sin x = \cos(x- \frac {\pi}2)$. So $\sin x$ is not in quadrature wirh $\cos(x- \frac {\pi}2)$.

You can use sites such as https://www.desmos.com/calculator to draw the graphs. Or use a graphing calculator or laptop+software or similar.

Sketch the graphs of $y=5\cos(11t + \frac {7\pi}2)$ and $y=2\sin(11t)$ for yourself. You will see the graphs are in phase, not in quadrature.

Generally, if the 'waves' do not have the same frequency and/or are not in quadrature, there is no shortcut formula/method I know of. Drawing the graph would be simplest/quickest method in this situation (IMO).

Ok, understand that both curves must be differ by 90 degree to make the $\(A^2)+(B^2)$ work. In this case, simply sum up the curves of function having different frequency would be the upper limit ?

 

[Steve4Physics]

Ok, understand that both curves must be differ by 90 degree to make the $\sqrt {A^2+B^2}$ work. [Edit - LateX corrected.] In this case, simply sum up the curves of function having different frequency would be the upper limit ?

I don't understand the question. If the frequencies are different you can't have a 90 degree phase difference - the phase difference continually changes.

The total amplitude can not be greater than the sum of the individual amplitudes - the sum is the upper limit.

If you are still unclear, post an example problem and what you think the answer is.

 

[FactChecker]

UPDATE: This is wrong. See the next post by @Steve4Physics for a much better analysis.

It might help to know that there are certain times when all the waves come back to their initial, t=0, values. For instance, at t=39,104 we have 39,104/13=3,008; 39,104/6.4=6,110; and 39,104/18.8=2080. So at that time, all the waves will start repeating. I believe this is the earliest time for them to all start repeating in unison.
I got this by factoring 130, 64, and 188 to get the Least Common Multiple (LCM), 195,520. Then I divided that by 10. At that time, all the angles are integer multiples of $\pi$, but one is not a multiple of $2\pi$. Multiplying the LCM by 2 gives a time, t=39,104 where all angles are integer multiples of $2\pi$.
At this time, it is not immediately clear to me what to do with this fact. I believe it would explain why I can not find an amplitude of 1475 or higher on a graph.

 

[Steve4Physics]

It might help to know that there are certain times when all the waves come back to their initial, t=0, values. For instance, at t=39,104 we have 39,104/13=3,008; 39,104/6.4=6,110; and 39,104/18.8=2080. So at that time, all the waves will start repeating. I believe this is the earliest time for them to all start repeating in unison.
I got this by factoring 130, 64, and 188 to get the Least Common Multiple (LCM), 195,520. Then I divided that by 10. At that time, all the angles are integer multiples of $\pi$, but one is not a multiple of $2\pi$. Multiplying the LCM by 2 gives a time, t=39,104 where all angles are integer multiples of $2\pi$.
At this time, it is not immediately clear to me what to do with this fact. I believe it would explain why I can not find an amplitude of 1475 or higher on a graph.

Consider two periodic continuous (not necessarily sinusoidal) real functions of time, $y_1$ and $y_2$, with respective periods $T_1$ and $T_2$.

The sum, $Y = y_1 + y_2$, has a period which is the least common multiple (LCM) of $T_1$ and $T_2$. That’s the shortest time-interval both functions will have simultaneously completed an exact integer number cycles.

So to find the amplitude of $Y$, it is necessary to find the maximum value during a time interval equal to LCM($T_1, T_2)$.

The argument extends to the sum of any number of periodic functions.
_________

Now consider the original problem. Note that the function $y = sin(\omega t)$, for example, has period $T = \frac {2\pi}{\omega}$.

In the Post #1 function, there are 3 different angular frequencies: $\omega_1 = 13\pi, \omega_2=6.4\pi$ and $\omega_3=18.8\pi$.

The corresponding 3 periods are therefore:
$T_1= \frac {2\pi}{13\pi} = 2/13$
$T_2= \frac {2\pi}{6.4\pi} = 2/6.4 = 20/64 = 5/16$
$T_3=\frac {2\pi}{18.8\pi} =2/18.8 = 20/188 = 5/47$

LCM($T_1, T_2, T_3$) = LCM (2/13, 5/16, 5/47) = 10, so the amplitude corresponds to the largest value within any time-interval of 10 (unknown time-units).

Carefully inspecting a graph covering t=0 to t=10, I found the maximum is 1479.173 at t=4.78611, though there are similar (slightly smaller) local maxima.

 

[Gavran]

It looks like the problem can only be solved numerically

I agree with the above.
In the case that a root mean square amplitude must be calculated, instead of a peak amplitude, the problem can be solved by using an analytical method.

 

[FactChecker]

I agree with the above.
In the case that a root mean square amplitude must be calculated, instead of a peak amplitude, the problem can be solved by using an analytical method.

I don't want to hijack this thread, but I wonder what analytical method can be used for the root mean square amplitude. I am very ignorant on this subject.

 

[Gavran]

I don't want to hijack this thread, but I wonder what analytical method can be used for the root mean square amplitude. I am very ignorant on this subject.

$$ f_{RMS}=\sqrt{\frac{1}{T}\int_{0}^{T}(f(t))^2dt} $$
In this case there will be:
$$ RMS_{Total}=\sqrt{RMS_1^2+RMS_2^2+RMS_3^2+RMS_4^2} $$
where
$\begin{align} RMS_1&=\sqrt{\frac{13}{2}\int_{0}^{\frac{2}{13}}(300\cos(13\pi t))^2dt}=\frac{300\sqrt2}{2}\nonumber\\ RMS_2&=\sqrt{\frac{6,4}{2}\int_{0}^{\frac{2}{6,4}}(500\sin(6,4\pi t))^2dt}=\frac{500\sqrt2}{2}\nonumber\\ RMS_3&=\sqrt{\frac{13}{2}\int_{0}^{\frac{2}{13}}(300\sin(13\pi t))^2dt}=\frac{300\sqrt2}{2}\nonumber\\ RMS_4&=\sqrt{\frac{18,8}{2}\int_{0}^{\frac{2}{18,8}}(600\cos(18,8\pi t))^2dt}=\frac{600\sqrt2}{2}\nonumber \end{align}$

 

[FactChecker]

$$ f_{RMS}=\sqrt{\frac{1}{T}\int_{0}^{T}(f(t))^2dt} $$
In this case there will be:
$$ RMS_{Total}=\sqrt{RMS_1^2+RMS_2^2+RMS_3^2+RMS_4^2} $$

But isn't that treating them as though the phase doesn't matter? What if you add two which are just the negative of each other so that the total signal is always 0?

 

[Steve4Physics]

But isn't that treating them as though the phase doesn't matter? What if you add two which are just the negative of each other so that the total signal is always 0?

My (non-mathematician’s) interpretation is this…

The Post #18 equation:
$RMS_{Total}=\sqrt{RMS_1^2+RMS_2^2+...}$
only applies under certain conditions:

From https://en.wikipedia.org/wiki/Root_mean_square

“Waveforms made by summing known simple waveforms have an RMS value that is the root of the sum of squares of the component RMS values, if the component waveforms are orthogonal (that is, if the average of the product of one simple waveform with another is zero for all pairs other than a waveform times itself).”​

Two waveforms which are the negative of each other, for example $\sin(at)$ and $\sin(-at)$, do not satisfy the orthogonality requirement.

So, if using the above equation, it’s necessary to check for mutual orthogonality between all pairs of waveforms.

 

[FactChecker]

I would expect something similar (or even related to) the variance of a sump of random variables:
If $Y = X_1+X_2+...+X_n, \text{ then } Var(Y)=\sum_{i=1}^n Var(X_i) + 2*\sum_{i\lt j} Cov(X_i, X_j)$

There seems to be some similarity of the concepts.

 

[Gavran]

It can be proved that a set of functions $$ \left\{\sin\left(\frac{n\pi x}{L}\right)\right\}_{n=1}^\infty\cup\left\{\cos\left(\frac{n\pi x}{L}\right)\right\}_{n=0}^\infty $$ is mutually orthogonal on $\left[-L,L\right]$ which makes the equation $$ RMS_{Total}=\sqrt{RMS_1^2+RMS_2^2+...+RMS_m^2} $$ applicable in this case.