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3 years ago

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When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 0.120 m. The mass is now displ

aced from its equilibrium position and undergoes simple harmonic oscillations. How long does it take the mass to complete one full oscillation?

1 answer:

AURORKA [14]3 years ago

6 0

Answer:

Time period of the oscillation will be 0.695 sec

Explanation:

We have given mass attached to the string m = 0.350 kg

The spring is stretched by 0.120 m

So x = 0.120 m

We know that force is given by F = Kx

So mg = kx

$0.350\times 9.8=k\times 0.120$

k = 28.5833

Now time period is given by

$T=2\pi \sqrt{\frac{m}{k}}=2\times 3.14\sqrt{\frac{0.35}{28.5833}}=0.695sec$

So time period of the oscillation will be 0.695 sec

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Answer:

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And replacing we have:

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

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

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:

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

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

$c = 3x10^8 m/s$

$\theta = 37$

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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