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MArishka [77]
3 years ago
9

For a solution to be a physical solution, it must satisfy several criteria. First, it must be continuous everywhere. Second, it

must have a continuous derivative everywhere, except for points where the potential energy becomes infinite (as it does at the walls of the box). Finally, it must be normalizable. In this case, you can check the first criterion by noting that the two functions Csin(nT/L) and 0 are continuous and that they have the same value at x = 0 and z = L, where their domains meet. To check the second criterion, simply take the first derivative of Csin(nTr/L) and 0. Both derivatives are continuous functions in their domains. The points where the domains meet are exactly the points where the potential energy becomes infinite, so you don't have to check for continuity there. The third criterion requires that there exist some value of C such that Since (z) is zero outside of the interval 0 < ¢ < L, this equation reduces to Use this equation to find the unique positive value of C. Express your answer in terms of L.

Physics
1 answer:
Zigmanuir [339]3 years ago
5 0

Answer:

Explanation:

solution

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A block–spring system vibrating on a frictionless, horizontal surface with an amplitude of 7.0 cm has an energy of 14 J. If the
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Answer:

E_T= 28J

Explanation:

The energy of Mass-Spring System the sum of the potential energy of the block plus the kinetic energy of the block:

E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} mv^2

Where:

\Delta x=Amplitude\hspace{3}or\hspace{3}d eformation\hspace{3} of\hspace{3} the\hspace{3} spring\\m=Mass\hspace{3}of\hspace{3}the\hspace{3}block\\k=Constant\hspace{3}of\hspace{3}the\hspace{3}spring\\v=Velocity\hspace{3}of\hspace{3}the\hspace{3}block

There are two cases, the first case is when the spring is compressed to its maximum value, in this case the value of the kinetic energy is zero, since there is no speed, so:

E_T=\frac{1}{2} k \Delta x^2\\\\14=\frac{1}{2} k7^2\\\\Solving\hspace{3} for\hspace{3} k\\\\k=\frac{28}{49} =\frac{4}{7}

The second case is when the block passes through its equilibrium position, in this case the elastic potential energy is zero since \Delta x=0, so:

E_T=\frac{1}{2} mv^2\\\\14=\frac{1}{2} mv^2\\\\Solving\hspace{3} for\hspace{3} v\\\\v^2=\frac{28}{m}

Now, let's find the energy of the system when the block is replaced by one whose mass is twice the mass of the original block using the previous data:

E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} m_2v^2

Where in this case:

m_2=New\hspace{3}mass=Twice\hspace{3} the\hspace{3} mass \hspace{3}of\hspace{3} the\hspace{3} original=2m

Therefore:

E_T=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{m_2})=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{2m})=14+14=28J

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