Title: Default for 4.5 Applications of Trigonometry Waves
Course: HELM Workbooks 2023-2024

d

Two-dimensional motion

Suppose that a wheel of radius \(R\) is rotating anticlockwise as shown in the figure below.

[Image: Fig 38]

Assume that the wheel is rotating with an angular velocity \(\omega\) radians per second about \(O\) so that, in a time \(t\) seconds, a point \((x,y)\) initially at position  \(A\) on the rim of the wheel moves to a position \(B\) such that angle \(AOB=\omega t\) radians.

Then the coordinates \((x,y)\) of  \(B\) are given by \[ \quad \qquad x=OP=R\cos\omega t\] \[ \quad \qquad y=OQ=PB=R\sin\omega t\]

We know that both the standard sine and cosine functions have period \(2\pi\). Since the angular velocity is \(\omega\) radians per second the wheel will make one complete revolution in \(\dfrac{2\pi}{\omega}\) seconds.

The time \(\dfrac{2\pi}{\omega}\) (measured in seconds in this case) for one complete revolution is called the period of rotation of the wheel.  The number of complete revolutions per second is thus \(\dfrac{1}{T}=f\) say which is called the frequency of revolution.  Clearly \(f =\dfrac{1}{T}=\dfrac{\omega}{2\pi}\) relates the three quantities introduced here.  The angular velocity \(\omega=2\pi f\) is sometimes called the angular frequency.

One-dimensional motion

The situation we have just outlined  is  two-dimensional motion.  More simply we might consider one-dimensional motion.

An example is the motion of the projection onto the \(x\)-axis of a point \(B\) which moves with uniform angular velocity \(\omega\) round a circle of radius \(R\) (see the figure below). As \(B\) moves round, its projection \(P\) moves to and fro across the diameter of the circle.

[Image: Fig 39]

The position of \(P\) is given by

\[x=R\cos \omega t\]

Clearly, from the known properties of the cosine function, we can deduce the following:

1. \(x\) varies periodically with \(t\) with period \(T=\dfrac{2\pi}{\omega}\).
2. \(x\) will have maximum value \(+R\) and minimum value \(-R\). (This quantity \(R\) is called the amplitude of the motion.)

Exercise

Using \(x=R\cos \omega t \) write down the values of \(x\) at the following times: \(t=0, \quad t=\dfrac{\pi}{2\omega}, \quad t=\dfrac{\pi}{\omega}, \quad t=\dfrac{3\pi}{2\omega}, \quad t=\dfrac{2\pi}{\omega}\).

\(t\) | 0 | \( \dfrac{\pi}{2\omega}\) | \( \dfrac{\pi}{\omega}\) | \( \dfrac{3\pi}{2\omega}\) | \( \dfrac{2\pi}{\omega}\)
\(x\) |  |  |  |  |
Feedback: Using \(x=R\cos \omega t \) we fill the table: \(t\) | 0 | \( \dfrac{\pi}{2\omega}\) | \( \dfrac{\pi}{\omega}\) | \( \dfrac{3\pi}{2\omega}\) | \( \dfrac{2\pi}{\omega}\) \(x\) | [value] | [value] | [value] | [value] | [value]

Using \(x = R \cos \omega t\) this 'to and fro' or 'vibrational' or 'oscillatory' motion between \(R\) and \(-R\)  continues indefinitely.  The technical name for this motion is  simple harmonic.  To a good approximation it is the motion exhibited (i) by the end of a pendulum pulled through a small angle and then released (ii) by the end of a hanging spring pulled down and then released. See the Figure below (in these cases damping of the pendulum or spring is ignored).

[Image: Fig40]

Using your knowledge of the cosine function and the results of the  previous  Task sketch the graph of \(x\) against \(t\) where \[ \quad x=R\cos \omega t   \text{ for } t=0 \text{ to } t=\dfrac{4\pi}{\omega}\]
Feedback: [Image: Task 1] This graph shows part of a cosine wave, specifically two periods of oscillation. The shape of the graph suggests that the term wave is indeed an appropriate description.

Exercise

We know that the shape of the cosine graph and the sine graph are identical but offset by \(\pi/2\) radians horizontally to the right. Bearing this in mind, attempt the following question.

Write the equation of the wave \(x(t)\), part of which is shown in the following graph. You will need to find the period \(T\), the angular frequency \(\omega\) and the amplitude \(A\) first.

Please write "pi" for the number \( \pi \).

[value]

\(T = \) \(\mathrm{s}\)

\(\omega = \) \(\mathrm{rad}/\mathrm{s}\)

\(A =\)

\( x = \)

The quantity \(x\), a function of \(t\), is referred to as the displacement of the wave.
Feedback: From the shape of the graph we have a sine wave rather than a cosine wave. The amplitude is \(A = [value]\). The period \(T = [value]\ \mathrm{s}\), so the angular frequency is \[\omega = \frac{2\pi}{T} = [value]\ \mathrm{rad}/\mathrm{s}.\] Hence, \[x = [value].\]

Time shifts between waves

We recall that \(\cos\left(\theta-\dfrac{\pi}{2}\right)=\sin\theta\) which means that the graph of \(x=\sin\theta\) is the same shape as that of \(x=\cos\theta\) but is shifted to the right by \(\dfrac{\pi}{2}\) radians.

Suppose now that we consider the waves \[ \quad \qquad x_1=R\cos 2t\qquad\qquad x_2=R\sin 2t\]

Both have amplitude \(R\), angular frequency \(\omega=2\) rad \(s^{-1}\). Also \[ \quad x_2=R\cos\left(2t-\dfrac{\pi}{2}\right)=R\cos\left[2\left(t-\dfrac{\pi}{4}\right)\right] \] The graphs of \(x_1\) against \(t\) and of \(x_2\) against \(t\) are said to have a time shift of \(\dfrac{\pi}{4}\). Specifically \(x_1\) is ahead of, or 'leads' \(x_2\) by a time \(\dfrac{\pi}{4}\) s.

More generally, consider the following two sine waves of the same amplitude and frequency: \[ \quad \qquad x_1(t)=R\sin\omega t\] \[ \quad \qquad x_2(t)=R\sin (\omega t-\alpha)\]

Now \(x_1\left(t-\dfrac{\alpha}{\omega}\right)=R\sin \left[\omega\left(t-\dfrac{\alpha}{\omega}\right)\right]=R\sin(\omega t-\alpha)=x_2(t)\) so it is clear that the waves \(x_1\) and \(x_2\) are shifted in time by \(\dfrac{\alpha}{\omega}\). Specifically \(x_1\) leads \(x_2\) by \(\dfrac{\alpha}{\omega}\) (if \(\alpha\ >0)\).

Exercise

Calculate the time shift between the waves

\[\begin{aligned} x_1 &= [value]\cos ([value] t) \\ x_2 &= [value]\cos \left([value] t +[value]\right) \end{aligned}\]

where the time \(t\) is in seconds.

\(\mathrm{s}\)
Feedback: We are given the waves \[\begin{aligned} x_1 &= [value]\cos ([value] t) \\ x_2 &= [value]\cos \left([value] t +[value]\right) \end{aligned}\] Note firstly that the waves have the same amplitude [value] and angular frequency \([value]\) (corresponding to a common period \(\dfrac{2\pi}{[value]}= \displaystyle{[value]} \text{ s}\)) . Now \(\cos \left([value] t+\displaystyle{[value]}\right) = \cos \left([value]\left( t +\displaystyle{[value]}\right)\right)\) so \(x_1\left( t +\displaystyle{[value]}\right)=x_2(t).\) In other words the time shift is \(\displaystyle{[value]}\ \text{s}\), the wave \(x_2\) leads the wave \(x_1\) by this amount. Alternatively we could say that \(x_1\) lags \(x_2\) by \(\displaystyle{[value]} s.\)

Time shift and phase difference

Key Point

The equations

\[ \quad x=R\cos\omega t\qquad\qquad x=R\sin\omega t \]

both represent waves of amplitude \(R\) and period \(\dfrac{2\pi}{\omega}\).

The time shift between these waves is \(\dfrac{\pi}{2\omega}\) because \(\cos \left\{\omega\left(t-\dfrac{\pi}{2\omega}\right)\right\}=\sin\omega t\).

The phase difference between these waves is said to be \(\dfrac{\pi}{2}\) because \(\cos {\left (\omega t-\dfrac{\pi}{2}\right)}=\sin {\omega t}\)

Combining two wave equations

A situation that arises in some applications is the need  to combine two trigonometric terms such as \[ \quad \qquad  A\cos\theta+B\sin\theta \quad \text{where } A \text{ and } B \text{ are constants} .\]

For example this sort of situation might arise if we wish to combine two waves of the same frequency but not necessarily the same amplitude or phase. In particular we wish to be able to deal with an expression of the form \[ \quad R_1\cos \omega t +R_2\sin\omega t\]  where the individual waves have, as we have seen, a time shift of \( \dfrac{\pi}{2\omega}\) or a phase difference of \( \dfrac{\pi}{2}\).

General Theory

Consider an expression \( A\cos\theta+B\sin\theta.\)  We seek to transform this into the single form \(C\cos (\theta-\alpha)\)  (or \(C\sin (\theta-\alpha))\), where \(C\) and \(\alpha\) have to be determined. The problem is easily solved with the aid of trigonometric identities.

We know that \[ \quad C\cos (\theta -\alpha)\equiv C(\cos\theta\cos\alpha+\sin\theta\sin\alpha)\]

Hence if \(A\cos\theta +B\sin\theta = C\cos (\theta -\alpha ) \) then\[ \quad A\cos\theta+B\sin\theta=(C\cos\alpha)\cos\theta+(C\sin\alpha)\sin\theta\]

For this to be an identity (true for all values of \(\theta\)) we must be able to equate the coefficients of \(\cos\theta\) and \(\sin\theta\) on each side.

Hence\[ \quad A=C\cos\alpha\qquad \text{and}\qquad B=C\sin \alpha\]

Exercise

1. By squaring and adding the equations \(A=C\cos\alpha\) and \(B=C\sin \alpha\), obtain \(C\) in terms of \(A\) and \(B\). You can assume \(C>0\).
\(C = \)
2. By eliminating \( C \) from the equations \(A=C\cos\alpha\) and \(B=C\sin \alpha\), obtain \(\tan\alpha \) in terms of \(A\) and \(B.\)
\(\tan\alpha = \)
Feedback: \(A= C\cos\alpha\) and \(B=C\sin\alpha\) gives \[ \quad A^2+B^2=C^2\cos^2\alpha+C^2\sin^2\alpha=C^2(\cos^2\alpha+\sin^2\alpha)=C^2\] therefore \( C=\sqrt{A^2+B^2}\) (We take the positive square root.) By division, \( \dfrac{B}{A}=\dfrac{C\sin\alpha}{C\cos\alpha}=\tan\alpha\). Furthermore, \(\alpha\) can be obtained by solving \(\tan\alpha=\dfrac{B}{A}\). However, care must be taken to obtain the correct quadrant for \(\alpha.\)

By eliminating \(C\) from Equations (2) and using the result of the previous Task, obtain \(\alpha\) in terms of \(A\) and \(B\).
Feedback: By division, \(\frac{B}{A}=\frac{C\sin\alpha}{C\cos\alpha}=\tan\alpha\) so \(\alpha\) is obtained by solving \(\tan\alpha=\frac{B}{A}\). However, care must be taken to obtain the correct quadrant for \(\alpha\).

Combining Wave Equations

Key Point

If  \(A\cos\theta+B\sin\theta= C\cos (\theta-\alpha)\)  then  \(C=\sqrt{A^2+B^2}\) and \( \tan\alpha=\dfrac{B}{A}.\)

Note that the following cases arise for the location of \(\alpha\):

 \[ \quad 1.\quad  A>0,  B>0 : \text{ 1st  quadrant } \] \[ \quad 2.\quad  A<0,  B>0 :\text{ 2nd quadrant } \] \[ \quad 3.\quad A<0, B<0 : \text{ 3rd quadrant } \] \[ \quad 4.\quad A>0, B<0 : \text{ 4th quadrant } \]

In terms of waves, using Key Point above we have\[ \quad R_1\cos\omega t+R_2\sin\omega t= R\cos (\omega t-\alpha)\]  where \(R=\sqrt{R^2_1+R^2_2} \)  and \(  \tan \alpha=\dfrac{R_2}{R_1}\).

The form \(R\cos (\omega t-\alpha)\) is said to be the  amplitude/phase form of the wave.

Example

Express in the form \(C\cos (\theta-\alpha)\) each of the following:

1. \(3\cos\theta+3\sin\theta\)
2. \(-3\cos\theta+3\sin\theta\)
3. \(-3\cos\theta-3\sin\theta\)
4. \(3\cos\theta-3\sin\theta\)

Solution

1. \(\tan\alpha=\dfrac{B}{A}=\dfrac{3}{3}=1\) gives \(\alpha=45^\circ\)   (\(A\) and \(B\) are both positive so the first quadrant is the correct one.) Hence \( 3\cos\theta+3\sin\theta=\sqrt{18}\cos(\theta-45^\circ)= \sqrt{18}\cos\left(\theta-\dfrac{\pi}{4}\right)\)
2. The angle \(\alpha\) must be in the second quadrant as \(A=-3<0\), \(B=+3>0\). By calculator : \(\tan\alpha=-1\) gives \(\alpha=-45^\circ\) but this is in the 4th quadrant. Remembering that \(\tan\alpha\) has period \(\pi\) or \(180^\circ\) we must therefore add \(180^\circ\) to the calculator value to obtain the correct \(\alpha\) value of \(135^\circ\). Hence \[ \quad \qquad -3\cos\theta+3\sin\theta=\sqrt{18}\cos(\theta-135^\circ)\]
3. Here \(A=-3,\ B=-3\) so \(\alpha\) must be in the 3rd quadrant. \(\tan \alpha=\dfrac{-3}{-3}=1\) giving \(\alpha=45^\circ\) by calculator. Hence adding \(180^\circ\) to this tells us that \[ \quad -3\cos\theta-3\sin\theta=\sqrt{18}\cos(\theta-225^\circ)\]
4. Here \(A=3,\ B=-3\) so \(\alpha\) is in the 4th quadrant. \(\tan\alpha=-1\) gives us (correctly) \(\alpha=-45^\circ\) so \[ \quad \qquad 3\cos\theta-3\sin\theta=\sqrt{18}\cos(\theta+45^\circ).\] Note that in the amplitude/phase form the angle may be expressed in degrees or radians.

Exercise

Given \( x= [value]\cos\omega t+[value]\sin \omega t\) write \(x\) in the form \(\alpha \cos(\omega t + \beta) \).

\(\alpha =\)

Using this value for \(\alpha\) we have

\( x =\)

Express the phase in radians to 3 d.p. and write "omega" for the parameter \( \omega\).
Feedback: We have \(x=R\cos(\omega t+\alpha)\) where \( R=\sqrt{[value]^2+[value]^2}=[value]\) and \( \tan\beta=\dfrac{[value]}{[value]}\) from which, using the calculator in radian mode, \(\beta=[value]\) radians. This is in the fourth quadrant \(\left(0>\beta >-\frac{\pi}{2}\right)\) which is correct since \(A=[value]\) is positive and \(B=[value]\) is negative. Hence, \[x=[value]\cos(\omega t+([value])).\]

Exercise

Write down the amplitude, R, and the period, T, of \( \displaystyle{y=[value]\sin\left( [value] t\right)} \).

Please write "pi" for the value \( \pi\).

\( \textrm{R} = \)

\( \textrm{T} = \)
Feedback: The wave equation \(y= \displaystyle{[value]\sin [value]t} \) has amplitude \(\textrm{R} = [value] .\) The period is \(\textrm{T} = \dfrac{2\pi}{[value]}=\displaystyle [value]\). Check: \(y(t+[value])=[value]\sin ([value](t+[value]))=[value]\sin ([value] t+2\pi)= [value]\sin [value] t=y(t)\).

Exercise

The depth \(y\) of water at the entrance to a small harbour at time \(t\ (\mathrm{h})\) is

\[ y=a\sin\left( b\left(t-\dfrac{\pi}{2}\right)\right)+k,\]

where \(k\ (\mathrm{m})\) is the average depth.

If the tidal period is \([value]\ \mathrm{h}\) and the depths at high tide and low tide are \( [value]\ \mathrm{m}\) and \( [value]\ \mathrm{m}\) respectively, obtain \(a,\ b,\ k\).

Please write "pi" for the value \( \pi\).

\(a=\) \(\mathrm{m}\)

\(b=\) \(\mathrm{rad}\,\mathrm{h}^{-1}\)

\(k=\) \(\mathrm{m}\)
Feedback: We have \(y=a\sin\left( b\left(t-\dfrac{\pi}{2}\right)\right)+k.\) Since the period is \(\dfrac{2\pi}{b}= [value] \text{ hr},\) we obtain \(b= \displaystyle [value]\ \mathrm{rad}\,\mathrm{h}^{-1}\). Furthermore, \[y_{\max}=a+k \quad \text{and} \quad y_{\min}=-a+k\] and hence \[a+k= [value] \quad \text{and} \quad -a+k=[value].\] If we sum these two equations, we obtain \(k=\displaystyle{[value]} \ \mathrm{m}\), and if we subtract them, we find \(a=\displaystyle{[value]} \ \mathrm{m}\). Therefore, the function that models the depth of the water is \[y=\displaystyle{[value] \sin \left([value]\left(t-\dfrac{\pi}{2}\right)\right)+[value]}.\]

Exercise

The Fahrenheit temperature at a certain location over one complete day is modelled by \[ \quad F(t)=[value]+[value]\sin\left( \dfrac{\pi}{[value]}(t-[value])\right)\qquad 0\le  t\le 24\] where \(t\) is time in hours after midnight.

1. What are the temperatures at \(8\) am and \(12\) noon?
Give your answer to \(1\) decimal place.
\( F(8) = \)

\( F(12) = \)
2. At what time(s) that day is the temperature \([value]\phantom{1}^\circ\mathrm{F}\)?
Write your answer as a set {t1,t2,...}.
\( t = \)
3. Obtain the maximum and minimum temperatures and the times at which they occur.
The maximum temperature is \(\phantom{1}^\circ\mathrm{F}\) which occurs at \(t_{max} = \)
The minimum temperature is \(\phantom{1}^\circ\mathrm{F}\) which occurs at \(t_{min} = \)
Feedback: 1. When \(t=8\), then the temperature is \([value]+[value]\sin \dfrac{\pi}{[value]}(8-[value])=[value]^\circ \,F.\) And for \(t=12\), we have \([value]+[value]\sin \dfrac{\pi}{[value]}(12-[value])=[value]^\circ \, F\). 2. \(F(t)=[value]\) when \( \dfrac{\pi}{[value]}(t-[value])=0,\ \pi,\ \ 2\pi,\ldots\). Therefore, \(t-[value]=0,[value],[value],\ldots\) and hence, \(t=[value],[value],[value],\ldots\) hours. Since \( t\le 24\), we obtain \(t=[value] \, ([value].00 \, \mathrm{am}) \) and \(t=[value] \, ([value].00 \, \mathrm{pm}).\) 3. Maximum temperature is \([value]^\circ\) F when \(\dfrac{\pi}{[value]}(t-[value])=\dfrac{\pi}{2}\) i.e. at \(t=[value]\). Minimum temperature is \([value]^\circ\)F when \(\dfrac{\pi}{[value]}(t-[value])=\dfrac{3\pi}{2}\) i.e. at \(t=[value]\).

Exercise

In each of the following write down expressions for time-shifted sine and time-shifted cosine functions that satisfy the given conditions:

1. Amplitude \( [value]\), Period \( \displaystyle[value]\), Time shift \(\displaystyle[value]\)
\( y = \) is the time-shifted sine function.
\( y = \) is the time-shifted cosine function.
2. Amplitude \( [value]\), Period \( [value]\), Time shift \( [value]\)
\( y =\) is the time-shifted sine function.
\( y =\) is the time-shifted cosine function.

Please write "pi" for the value \( \pi\).
Feedback: 1. Amplitude \( [value]\), Period \( [value]\), Time shift \( [value]\) so: \( y=[value] , \quad y=[value].\) 2. Amplitude \( [value]\), Period \( [value]\), Time shift \( [value]\) so: \( y=[value] , \quad y=[value].\)

Exercise

Express the sum of  two a.c. currents

\[ i_1=[value] \quad \text{and} \quad i_2=[value],\]

in the form \(\displaystyle{A\cos\left(\omega t + \alpha\right)} \).

\( i_1+i_2= \)

Please write "pi" for the value \( \pi\) and "omega" for the parameter \( \omega\).
Feedback: We are given the two a.c. currents \[ i_1=[value] \quad \text{and} \quad i_2=[value].\] Since \( \sin x=\cos\left(x-\dfrac{\pi}{2}\right)\) we have: \( \sin\left(\omega t+[value]\right)=\cos\left(\omega t+[value]-\dfrac{\pi}{2}\right)= \cos \left(\omega t [value]\right). \) Therefore \( i_1+i_2=[value] + [value]=[value]\)

Exercise

The motion of a weight on a spring is given by \[\displaystyle{y= [value]\cos([value]t)- [value]\sin([value]t)}\]

Obtain \(C\) and \(\alpha\) such that \( \displaystyle{y=C\sin ([value]t+\alpha)}\).

\( y =\)
Feedback: Using the formulae gives: \[ C=\sqrt(([value])^2+ ([value])^2)=[value] \] \[\cos\alpha=-\dfrac{[value]}{[value]} \qquad \sin \alpha = -\dfrac{[value]}{[value]}\] We see that \(\alpha\) is in the second quadrant and \(\alpha=[value]\). Therefore, \[y=[value] \]

Exercise

Given that if \( A\cos\omega t+B\sin\omega t= C\sin (\omega t+\alpha)\)  then \[\quad C=\sqrt{A^2+B^2},\qquad \ \cos\alpha=\dfrac{B}{C}, \qquad \sin\alpha=\dfrac{A}{C}.\]

Express the following in the amplitude/phase form \( C\sin (\omega t+\alpha). \) Approximate all floating point numbers up to \(4\) d.p.

1. \(y= [value]\cos([value]t)- [value]\sin([value]t), \)
\( y = \)
2. \(y= [value]\cos([value]t)+ [value]\sin([value]t), \)
\( y = \)
Feedback: Since \[\sin(\omega t+\alpha)=\sin\omega t\cos\alpha+\cos\omega t\sin \alpha\] then \(A=C\sin\alpha\) (coefficients of \(\cos\omega t\)) and \(B=C\cos\alpha\) (coefficients of \(\sin\omega t\),) from which \[C^2=A^2+B^2,\quad \sin\alpha =\dfrac{A}{C},\quad \cos \alpha=\dfrac{B}{C}.\] 1. \( C=\sqrt{[value]^2+ [value]^2}, \) \( \cos\alpha=-\dfrac{[value]}{[value]}\) and \( \sin \alpha-\dfrac{[value]}{[value]},\) so \(\alpha\) is in the second quadrant and \(\alpha=[value].\) Therefore \(y=[value] \). 2. Similarly, \(y=[value] \).

Exercise

Write the a.c. current \( i=[value]\cos([value]t)+ [value]\sin([value]t) \) in the form \( C\cos([value]t-\alpha) \).

Approximate all floating point numbers up to \(4\) d.p.

\(i = \)
Feedback: We are given the a.c. current \( i=[value]\cos([value]t)+ [value]\sin([value]t) .\) We know that \(C= \sqrt{[value]^2+ [value]^2}\). \( \tan\alpha=\dfrac{[value]}{[value]}\) and \(\alpha\) must be in the first quadrant (since \(A=[value], B=[value] \) are both positive.) Therefore \( \alpha=\tan^{-1}\dfrac{[value]}{[value]}=[value] \, \mathrm{rad}\), and then \[i=[value].\]

Exercise

The current in an a.c. circuit is

\[ i(t)=[value] \sin ([value]t)\ \mathrm{amp},\]

where \(t\) is measured in seconds.

What is the maximum current and at what times does it occur?

Please write "pi" for the value \(\pi\).

The maximum current is \(\mathrm{amp}\),

which occurs at times \( t = \) \(\mathrm{s}\), for \(n=1,2,3 \dots \)
Feedback: The current in an a.c. circuit is \[ i(t)=[value] \sin ([value]t)\ \mathrm{amp}.\] The maximum value of the sine function is 1, therefore the maximum value is \(i(t) = [value] \, \mathrm{amp}\). Then to find the value of \(t\) for which this maximum occurs, find \(\sin ([value]t) = 1\). Firstly \([value]t = [value]\) for \( n \in \mathbb{Z}\). Hence, \[t = [value] = [value] = [value].\]

Exercise

Calculate the amplitude \(R\), frequency \(f\) and time shift \(\alpha\) between the waves.

1. \( x_1=[value] \quad \text{and} \quad \displaystyle{x_2 =[value]} \)
\( R = \)
\( f = \)
\( \alpha = \)
2. \( \displaystyle{y_1 = [value] \quad \text{and} \quad y_2=[value]} \)
\( R = \)
\( f = \)
\( \alpha =\)

Please write "pi" for the value \( \pi\).
Feedback: 1. We are given the waves \( x_1=[value] \quad \text{and} \quad \displaystyle{x_2 =[value]} \) The amplitude is [value]. The angular frequency is \( \omega = [value] \). So the frequency is \( f = \dfrac{\omega}{2\pi} = \displaystyle [value].\) Writing \(\displaystyle{ x_2 = [value]\sin \left[ [value] \left(t-[value]\right) \right] }\) we see that there is a time shift of \(\displaystyle{[value]}\) in this wave compared with \(\displaystyle{ x_1= [value] }.\) 2. We are given the waves \( \displaystyle{y_1 = [value] \quad \text{and} \quad y_2=[value]}. \) The amplitude is [value]. The angular frequency is \( \omega = [value] \). So the frequency is \( f = \dfrac{\omega}{2\pi} = \displaystyle [value].\). Writing \(\displaystyle{ y_2=[value]\cos\left[ [value] \left(t-[value]\right) \right] }\) we see that there is a time shift of \(\displaystyle{[value]} \) in this wave compared with \(\displaystyle{y_1=[value]} \).

Exercise

Using the values you filed in on the previous table, what is the shape of the function \( x = R\cos(\omega t) \) for \( t=0\) to \( t=\dfrac{4\pi}{\omega}\), where \( R=[value]\) and \( \omega=[value]  \)?

Choose from the graphs below:

Graph A | Graph B
[value] | [value]
Graph C | Graph D
[value] | [value]
Feedback: Here are the values we found for \(x = R \cos(\omega t)\) in the previous question. \(t\) | \( 0 \) | \( \dfrac{\pi}{2\omega} \) | \( \dfrac{\pi}{\omega} \) | \( \dfrac{3\pi}{2\omega} \) | \( \dfrac{2\pi}{\omega} \) \(x\) | \(R\) | \(0\) | \(-R\) | \(0\) | \(R\) Substituing the given values for \(R\) and \(\omega\) we see that the correct graph is [value]. [value]