- #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})##

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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

##\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.

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.