Superposition of two one-dimensional harmonic waves

In summary: I suggested in post #11.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 and the maximum amplitude, as I suggested in post #11.Oh I see. So for part c) the answer would be ##x_{max_{2}}=0## and ##A_{max} = 7cm##?In summary, the superposition of two one-dimensional harmonic waves with equations ##s_1(x,t) = 3.5 cm \cdot cos(27.5s^{-1} \cdot t - 5.65m^{-1} \cdot x)## and ##s
  • #1
orangephysik
11
1
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})##

--------------------------------------------------------------------------------------------------------------
##\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.
 
Physics news on Phys.org
  • #2
Your part a answers look fine to me.
In part b,
orangephysik said:
##\cos(0.075)##
Did you forget something ?
 
  • #3
haruspex said:
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)
 
  • #4
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.
 
  • #5
nasu said:
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?
 
  • #6
orangephysik said:
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.
 
  • Like
Likes orangephysik and nasu
  • #7
orangephysik said:
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.
 
  • Like
Likes orangephysik
  • #8
Oh okay. Thank you :smile:
 
  • Like
Likes hutchphd
  • #9
nasu said:
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?"
 
Last edited:
  • #10
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
1679345518635.png

for ##\cos##
 
  • Like
Likes orangephysik
  • #11
haruspex said:
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.
 

Related to Superposition of two one-dimensional harmonic waves

What is the superposition principle in the context of harmonic waves?

The superposition principle states that when two or more harmonic waves overlap in space, the resultant displacement at any point is the algebraic sum of the displacements due to each individual wave. This principle applies to all linear systems, including one-dimensional harmonic waves.

How do you mathematically represent the superposition of two one-dimensional harmonic waves?

Mathematically, if you have two harmonic waves represented as \( y_1(x, t) = A_1 \sin(k_1 x - \omega_1 t + \phi_1) \) and \( y_2(x, t) = A_2 \sin(k_2 x - \omega_2 t + \phi_2) \), their superposition can be expressed as \( y(x, t) = y_1(x, t) + y_2(x, t) \). This results in a new wave that combines the properties of both original waves.

What is the physical significance of the superposition of two harmonic waves?

The physical significance of the superposition of two harmonic waves can manifest as constructive or destructive interference. Constructive interference occurs when the waves are in phase, leading to an increased amplitude, while destructive interference occurs when the waves are out of phase, leading to a decreased amplitude or even cancellation.

What are beats in the context of superposition of harmonic waves?

Beats are a phenomenon that occurs when two harmonic waves of slightly different frequencies interfere with each other. The result is a new wave pattern that oscillates in amplitude at a frequency equal to the difference between the two original frequencies. This causes a periodic variation in loudness known as the beat frequency.

How does the superposition of two harmonic waves relate to standing waves?

Standing waves are a special case of wave superposition where two waves of the same frequency and amplitude travel in opposite directions. The superposition of these waves results in a wave pattern that appears to be stationary, characterized by nodes (points of no displacement) and antinodes (points of maximum displacement). This occurs due to the constructive and destructive interference of the two waves at fixed points in space.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
782
  • Introductory Physics Homework Help
Replies
3
Views
877
  • Introductory Physics Homework Help
Replies
10
Views
430
Replies
9
Views
145
  • Introductory Physics Homework Help
Replies
16
Views
542
  • Introductory Physics Homework Help
Replies
9
Views
760
  • Introductory Physics Homework Help
Replies
4
Views
353
  • Introductory Physics Homework Help
Replies
6
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
352
  • Introductory Physics Homework Help
Replies
3
Views
270
Back
Top