Write a sinusoidal equation, y=a sin(bx+c)+d, with the characteristics given.?

Question

Maximum=5
Period=7
Start Point=(1.1,4)

It is a question in one of my Applied Math 30 modules, worth 5 marks, so Im just assuming there has to be some way to do it

Answer 1

The wave equation is typically – for time t

y(t) = Asin(ωt + φ) + b

Where

A is the amplitude – the distance from the time-axis (y=0) and the peak of the wave
ω is the angular frequency = 2π/T = 2π f where T is the period – the time for one complete cycle, and f is the frequency – the number of periods per unit time
φ is the phase shift – it shifts the curve along the time-axis (i.e. to the right)
b is the bias – it shifts everything up the y-axis

Now here x is used instead of t so

y(x) = Asin(2πfx + φ) + b = asin(bx+c)+d

And a = A, b = 2πf, c= φ and d = b

Now from this the maximum = a+d so a+d=5

the period = T = 1/f so f = 1/7, so b = 2π/7 = 0.8976

Starting point is x=1.1 and y=4

Now y = 4 is the bias so d = 4, and a = 1

And x=1.1 is the phase i.e. sin(bx+c) = 0 when x = 1.1

So as b = 0.9 c = -0.8976*1.1 = -0.9874

So to sum up

y = sin(0.8976x - 0.9874) + 4

Another way to solve this is to just treat it just like any equation with boundary conditions:

So if, y=a sin(bx+c)+d

Then y' = abcos(bx+c) and y'' = -ab²sin(bx+c)

And to find the value of x where a y is a maximum - find y' = 0 where y" = -ve

So abcos(bx+c) = 0 and so bx+c = ±(2k+1)π/2 where k = 0,1,...

However, we only need to look at the first solution

where bx+c = π/2

And so ymaximum = asin(π/2)+d = a + d

So a + d = 5

The period of a function is D where f(x) = f(x+D) for all values of x - i.e. the function repeats

Clearly here as we know sin repeats every 2π

b(x+D) + c = bx + c + 2π

So b(x+D) - bx = 2π

and bD = 2π

so b = 2π/D, and as D = Period = 7

b = 2π/7 = 0.8976

So far we have:

a+d = 5 and b = 2π/7

The "Start Point" is the point where the curve "starts", Normally it woud be at y = 0 - but the factor d shifts it up to y=4

So clearly d = 4, so a = 1

and y = sin(2π/7x + c) + 4 = 4 so

sin((2π/7)1.1 + c) = 0

and so c = -(2π/7)*1.1 = -0.9874

Which gives the same answers as before

Answer 2

You have 4 constants(a, b, c, d) to be evaluated. You need 4 conditions whereas givens = 3 conditions (insufficient to solve). Interesting point is start point as I understood as y = 4 when x = 1.1. Even without mentioning start point, this is a simple condition only - nothing special.

y = a sin(bx+c) + d

Ymax = a + d = 5 when bx + c = (pi)/2
a + d = 5 .... ..... ..... (1)

2(pi) / T = b where T = Period = 7
2(pi) / 7 = b
b = 2(pi) / 7 ..... ..... (2)

If I read correctly, (1.1, 4)
y = 4, when x = 1.1
Plugging
4 = a sin(1.1b + c) + d
Plugging value of b
4 = a sin[1.1 x 2(pi)/7 + c] + d
4 = a sin[0.987 + c] + d
Plugging d = 5 - a
4 = a sin[0.987 + c] + 5 - a
a - a sin[0.987 + c] = 1
a [1 - sin(0.987 + c)] = 1
Here we need 1 more condition to work out two values
For this problem, assume a = 1 and you get d = 5 - 1 = 4
1 - sin(0.987 + c) = 1
sin(0.987 + c) = 0
0.987 + c = 0 [first quadrant value]
c = - 0.987

So the assumed 1 solution (which may be varied ]
y = sin[(2 pi/7)x - 0.987] + 4

Another solution
Assume a = 2, then d = 5 - 2 = 3
2 [1 - sin(0.987 + c)] = 1
2 sin(0.987 + c) = 2 - 1 = 1
sin(0.987 + c) = 1/2 = sin(pi/6) [first quadrant value]
0.987 + c = pi/6 = 0.524
c = - 0.463

Solution 2 with assumed value of a = 2
y = 2 sin[(2 pi/7)x - 0.463] + 3

You may continue for further solutions too.
Enjoy your time

Answer 3

Personally, I prefer the form y = a sin[b(x - c)] +d, because then the variable a is the amplitude, b tells us the period (the period will be 2(pi) divided by b), and c and d are the horizontal and vertical translations respectively.

The question is poorly phrased, but we can answer if we make a couple of assumptions (which like all assumptions, may or may not be valid).

First, we are told that 5 is the maximum, but we do not know what the minimum is, so we cannot know the amplitude. If we assume that the questioner meant that the maximum is 5 and the minimum is -5, then a=5.

Secondly, what is meant by "Start point" is not clear, since sinusoidal curves extend infinitely in both directions. But if we assume that what is meant is that the curve passes through that point and that the point (1, 1.4) is the image of the origin (0, 0)under a translation, then we can proceed. d will be 1.4, the vertical translation, and 1 is the horizontal translation.

Finally in this case, we are given that the period is 7 so b is 7 over 2(pi).

So we have y = 5 sin [(7/(2pi))(x - 1)] + 1.4
or y = 5 sin (7x/2pi - 7/2pi) + 1.4

Answer 4

To do this problem, you must have a thorough grounding on the wave structure and the parameters which define them. Sinudoisal waves are defined by Amplitude, wave length and the origin (and phase difference).

Now the generic equation for Sine waves can be derived as follows.

* * * * * *
Amplitude = Sin (degree in Radian). However the amplitude here would vary between 1 and -1. To have a greater amplitute than this you must have a amplitude multiplier say "A". Then the equation becomes Amplitude = A Sin (Degree in Radian). Now the amplitude will vary between -A to + A.

* * * * * *
Time Period(T): The time period is the time taken for the wave to complete 1 cycle. Sine waves default period is 360 degree or 2π, however this can be altered by multiplying a "Time Period multiplier "B". To attain a time period T, the multiplier B would be 2π / T.

* * * * * *
Phase difference: The default wave passes through the origin, however to move the starting point on the X axis, you need to add the value "C" to the degree.

Since this will alter the starting point of the wave on the x-axis, we can get the value by multiplying the "shift in the starting point with 2π / T". So C = starting point in X axis * 2π / T. To shift the starting point in +ve X direction, we need to substract the value and vice-versa.

Thus the generic wave equation would be Amplitude = A Sin (Degree * B + C).

* * * * * *
Axis of wave motion: The axis of the wave is always taken as Y = 0. However, you can change the axis by adding an Axis transformer "D". So the generic equation becomes Amplitude = D + A Sin (Degree * B + C).
* * * * * *

So in the given equation [ y=a sin(bx+c)+d ] each variable represent:

y= Amplitude
x= degrees in radians
a= Amplitude multiplier
b= Angular frequency multiplier
c= Phase difference (To shift the starting point on X axis)
d= X Axis transformer.

If I put the values you have given to the mentioned parameters, we get something like:

a= (5-4 = 1) as this is the maximum the wave can oscillate. Since the wave is oscillating on the axis Y = 4, the amplitude need to be only 1, to attain maximum value of 5.

b= Time period= Since the default time period is 2π. To get the time period as 7, the value of b would be ((2π) / 7)

c= Phase difference = Since we need a phase difference of 1.1, we need to multiply the fraction (1.1 / 7) * 2pi. Hence the value of c would be - 1.1*(2π/7)).

d= Axis transformer = 4, as given in the starting point.

Hence the final equation would be

y = 1 * sin (x*(2π/7) - 1.1*(2π/7)) + 4.

You can plot this graph and check if this is what you wanted.

Answer 5

upward thrust over run... upward thrust of -6 over a run of -3...for this reason your line has a slope of two. Now, you want to understand an equation for a line on a graph is: y = mx + b A parallel line could have a similar slope because the line you've been given. If m (slope) = 2 and your x-intercept (b) = a million/3 then: y = (2)x + a million/3

Answer 6

impossible with those given