Superposition of two one-dimensional harmonic waves

[orangephysik]

Homework Statement

(See below)

Relevant Equations

(See below)

$\mathbf {Homework ~Statement:}$
Consider the superposition of two one-dimensional harmonic waves
$$s_1(x,t)=3.5 cm \cdot cos(27.5s^{-1} \cdot t - 5.65m^{-1} \cdot x)$$
$$s_2(x,t)=3.5 cm \cdot cos(27.5s^{-1} \cdot t - 5.5m^{-1} \cdot x)$$

$\mathbf {a)}$ Calculate the wavelength $\lambda$, the propagation speed $v$ and the period $T$ for both waves
$\mathbf {b)}$ Calculate the superposition $s(x,t)$ of both waves
$\mathbf {c)}$ For which $x_{max}$ will the amplitude be a maximum? What are these values?

$\mathbf {Relevant ~Equation:}$
$cos(\alpha)+cos(\beta) = 2 \cdot cos(\frac{\alpha + \beta}{2}) \cdot cos(\frac{\alpha - \beta}{2})$

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$\mathbf {Attempt ~at~ a ~Solution:}$
$\mathbf {a)}$ Well the equations are in the form of $u(x,t) = a \cdot cos(\omega t \mp kx)$,
whereby $|k| = \frac{2\pi}{\lambda} = \frac{\omega}{v}$
and $\omega = 2\pi f=\frac{2\pi}{T}$

I get

	$\lambda$	$v$	$T$
$s_1(x,t)$	$1.11m$	$4.86 m/s$	$0.228 s$
$s_2(x,t)$	$1.14m$	$4.99 m/s$	$0.228 s$

$\mathbf {b)}$ Using the relevant equation I got
$s_1 + s_2 =$ $7 cm \cdot cos(\frac{55s^{-1}\cdot t-11.15 m^{-1}\cdot x}{2})cos(0.075)$

$\mathbf {c)}$ I considered the case for $cos(0)=cos(\pi)=1$
and got
$x=\frac{55s^{-1}\cdot t}{11.15 m^{-1}}$
and
$x=-\frac{2\pi - 55s^{-1}\cdot t}{11.15 m^{-1}}$
and so the maximum amplitude would then be
$x_{max}=\pm 7cm\cdot cos(0.075)$

Are my solutions correct? I remember for part (a) I got 0 points in the exam. I don't know what I did wrong.

 

[haruspex]

Your part a answers look fine to me.
In part b,

$\cos(0.075)$

Did you forget something ?

 

[orangephysik]

Your part a answers look fine to me.
In part b,

Did you forget something ?

Oh right, that should be $cos(0.075 m^{-1} \cdot x)$ . I forgot the units.
So $s_{1} + s_{2} =7 cm \cdot cos(\frac{55s^{-1}\cdot t-11.15 m^{-1}\cdot x}{2})cos(0.075m^{-1} \cdot x)$
and so the maximum would be at when $cos(0)=1$. I get $x_{max_{1}} = \frac{55s^{-1}\cdot t}{11.15 m}$ and $x_{max_{2}} =0$
which means the values of the maximum amplitude would then be
$7 cm \cdot cos(0.37s^{-1} \cdot t)$ for $x_{max_{1}}$
$7 cm \cdot cos(27.5s^{-1}\cdot t)$ for $x_{max_{2}}$
Since the second value is greater, then only. $x_{max_{2}} =0$ ?
(I omitted the case for $cos(\pi)=-1$ since this would give amplitudes of the same magnitude, but only with a minus sign at the front)

 

[nasu]

The amplitude is not time dependent. It should be a fixed value for a given position, x.
Everything that multiplies the time dependent term is the amplitude.

 

[orangephysik]

The amplitude is not time dependent. It should be a fixed value for a given position, x.
Everything that multiplies the time dependent term is the amplitude.

Oh right.

So
$s_{1} + s_{2} =7 cm \cdot cos(\frac{55s^{-1}\cdot t-11.15 m^{-1}\cdot x}{2})cos(0.075m^{-1} \cdot x)$
⇔$s_{1} + s_{2} =7 cm \cdot cos(27.5s^{-1}\cdot t-5.575 m^{-1}\cdot x)cos(0.075m^{-1} \cdot x)$

With $cos(\alpha - \beta)=cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta)$:

⇔$s_{1} + s_{2} =7 cm \cdot [ (cos(27.5s^{-1}\cdot t) cos(-5.575 m^{-1}\cdot x) +sin(27.5s^{-1}\cdot t)sin(-5.575 m^{-1}\cdot x) )~~cos(0.075m^{-1} \cdot x) ]$
⇔$s_{1} + s_{2} =7 cm \cdot [ cos(27.5s^{-1}\cdot t) cos(-5.575 m^{-1} \cdot x)cos(0.075m^{-1} \cdot x) +sin(27.5s^{-1}\cdot t)sin(-5.575 m^{-1}\cdot x)cos(0.075m^{-1} \cdot x) ]$

Leaving out the terms with t, the amplitude is

$A=s_{1} + s_{2} =7 cm \cdot [ cos(-5.575 m^{-1} \cdot x)cos(0.075m^{-1} \cdot x) +sin(-5.575 m^{-1})cos(0.075m^{-1} \cdot x) ]$

with $x_{max}=0$

the maximum amplitude is $A_{max}=7cm \cdot (1+0) = 7cm$

This should be correct?

 

[vela]

Oh right.

So
$s_{1} + s_{2} =7 cm \cdot cos(\frac{55s^{-1}\cdot t-11.15 m^{-1}\cdot x}{2})cos(0.075m^{-1} \cdot x)$
⇔$s_{1} + s_{2} =7 cm \cdot cos(27.5s^{-1}\cdot t-5.575 m^{-1}\cdot x)cos(0.075m^{-1} \cdot x)$

With $cos(\alpha - \beta)=cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta)$:

⇔$s_{1} + s_{2} =7 cm \cdot [ (cos(27.5s^{-1}\cdot t) cos(-5.575 m^{-1}\cdot x) +sin(27.5s^{-1}\cdot t)sin(-5.575 m^{-1}\cdot x) )~~cos(0.075m^{-1} \cdot x) ]$
⇔$s_{1} + s_{2} =7 cm \cdot [ cos(27.5s^{-1}\cdot t) cos(-5.575 m^{-1} \cdot x)cos(0.075m^{-1} \cdot x) +sin(27.5s^{-1}\cdot t)sin(-5.575 m^{-1}\cdot x)cos(0.075m^{-1} \cdot x) ]$

Leaving out the terms with t, the amplitude is

$A=s_{1} + s_{2} =7 cm \cdot [ cos(-5.575 m^{-1} \cdot x)cos(0.075m^{-1} \cdot x) +sin(-5.575 m^{-1})cos(0.075m^{-1} \cdot x) ]$

with $x_{max}=0$

the maximum amplitude is $A_{max}=7cm \cdot (1+0) = 7cm$

This should be correct?

It would be simpler and less work for you if you identified the amplitude from
$$s_{1} + s_{2} =(7~{\rm cm}) \cos(27.5~{\rm s^{-1}} \cdot t - 5.575~{\rm m^{-1}} \cdot x)\cos(0.075~{\rm m^{-1}} \cdot x)$$ since the time dependence is isolated to one factor.

 

[nasu]

Oh right.

So
$s_{1} + s_{2} =7 cm \cdot cos(\frac{55s^{-1}\cdot t-11.15 m^{-1}\cdot x}{2})cos(0.075m^{-1} \cdot x)$
⇔$s_{1} + s_{2} =7 cm \cdot cos(27.5s^{-1}\cdot t-5.575 m^{-1}\cdot x)cos(0.075m^{-1} \cdot x)$

With $cos(\alpha - \beta)=cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta)$:

⇔$s_{1} + s_{2} =7 cm \cdot [ (cos(27.5s^{-1}\cdot t) cos(-5.575 m^{-1}\cdot x) +sin(27.5s^{-1}\cdot t)sin(-5.575 m^{-1}\cdot x) )~~cos(0.075m^{-1} \cdot x) ]$
⇔$s_{1} + s_{2} =7 cm \cdot [ cos(27.5s^{-1}\cdot t) cos(-5.575 m^{-1} \cdot x)cos(0.075m^{-1} \cdot x) +sin(27.5s^{-1}\cdot t)sin(-5.575 m^{-1}\cdot x)cos(0.075m^{-1} \cdot x) ]$

Leaving out the terms with t, the amplitude is

$A=s_{1} + s_{2} =7 cm \cdot [ cos(-5.575 m^{-1} \cdot x)cos(0.075m^{-1} \cdot x) +sin(-5.575 m^{-1})cos(0.075m^{-1} \cdot x) ]$

with $x_{max}=0$

the maximum amplitude is $A_{max}=7cm \cdot (1+0) = 7cm$

This should be correct?

You don't need all this. You already had the formula for the sum in post #1. All you had to do was to identify the amplitude.

 

[orangephysik]

Oh okay. Thank you

 

[haruspex]

You don't need all this. You already had the formula for the sum in post #1. All you had to do was to identify the amplitude.

No, the question asks for a value of x which maximises the amplitude. In the formula obtained in post #1, because the x factor had been omitted from the cos argument, it seemed like the amplitude was independent of x.
It is unclear whether the value of the maximum amplitude is even required in the answer. $x_{max}$ has an infinite sequence of values, and that may be what is meant in "What are these values?"

 

[malawi_glenn]

You do not need to type $m^{-1}$ and $s^{-1}$ all the time, just write that the units of $x$ is meters and $t$ is seconds. You can also write that $s_1$ and $s_2$ are in cm.

Also write

for $\cos$

 

[nasu]

No, the question asks for a value of x which maximises the amplitude. In the formula obtained in post #1, because the x factor had been omitted from the cos argument, it seemed like the amplitude was independent of x.
It is unclear whether the value of the maximum amplitude is even required in the answer. $x_{max}$ has an infinite sequence of values, and that may be what is meant in "What are these values?"

You are right, he got the right form in post #3, not #1.
From this one he can find the positions of maximum amplitude.