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2 years ago

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Consider the waveform expression. y (x, t) = ym sin (0.333x + 5.36 + 585t) The transverse displacement (y) of a wave is given as

a function of position (x in meters) and time (t in seconds) by the expression. Determine the wavelength, frequency, period, and phase constant of this waveform meters a. Metersb. Hertz c. seconds d. radians

1 answer:

Sonja [21]2 years ago

8 0

Explanation:

The waveform expression is given by :

$y(x,t)=y_m\ sin(0.333x+5.36+585t)$ ...........(1)

Where

y is the position

t is the time in seconds

The general waveform equation is given by :

$y(x,t)=y_m\ sin(kx+\phi+\omega t)$ ..........(2)

Where

$k=\dfrac{2\pi}{\lambda}$

$\omega=2\pi f$

On comparing equation (1) and (2) we get :

$0.333=\dfrac{2\pi}{\lambda}$

$\lambda=18.86\ m$

$585=2\pi f$

f = 93.10 Hz

Time period, $T=\dfrac{1}{f}$

$T=\dfrac{1}{0.010}$

T = 0.010 s

Phase constant, $\phi=5.36\ radian$

Hence, this is the required solution.

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Explanation

Part a

For this case we can use the Compton shift equation given by:

$\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

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So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

Part b

For this cas we can calculate the wavelength of the phton with this formula:

$\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

$E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

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