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SOVA2 [1]
2 years ago
12

14. If the spring constant of a simple harmonic oscillator is doubled, by what factor will the mass of the system need to change

in order for the frequency of the motion to remain the same?
Physics
1 answer:
klio [65]2 years ago
8 0

Lets se

And

\\ \rm\Rrightarrow T=2\pi\sqrt{\dfrac{m}{k}}

\\ \rm\Rrightarrow \sqrt{k}T=2\pi\sqrt{m}

So

\\ \rm\Rrightarrow k\propto m

If spring constant is doubled mass must be doubled

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1. 1. are the charged parts of an atom.
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The child's linear speed is
              
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Compute the kinetic energy of a proton (mass 1.67×10−27kg ) using both the nonrelativistic and relativistic expressions for spee
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The non-relativistic kinetic energy of a proton is 6.76\times10^{-12}\ J

The relativistic kinetic energy of a proton is 7.25\times10^{-12}\ m/s

Explanation:

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Mass of proton m=1.67\times10^{-27}\ kg

Speed v= 9.00\times10^{7}\ m/s

We need to calculate the kinetic energy for non relativistic

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

Put the value into the formula

K.E=\dfrac{1}{2}\times1.67\times10^{-27}\times(9.00\times10^{7})^2

K.E=6.76\times10^{-12}\ J

We need to calculate the kinetic energy for relativistic

Using formula of kinetic energy

K.E=mc^2(\sqrt{(\dfrac{1}{1-\dfrac{v^2}{c^2}})}-1)

K.E=1.67\times10^{-27}\times(3\times10^{8})^{2}\cdot\left(\sqrt{\frac{1}{1-\frac{\left(9.00\times10^{7}\right)^{2}}{(3\times10^{8})^{2}}}}-1\right)

K.E=7.25\times10^{-12}\ m/s

Hence, The non-relativistic kinetic energy of a proton is 6.76\times10^{-12}\ J

The relativistic kinetic energy of a proton is 7.25\times10^{-12}\ m/s

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3 years ago
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7 0
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