Respuesta :
Answer:
Explanation:
Assuming motion of equation of SHM as
[tex]x=A\cos \omega t[/tex]
where, [tex]A=amplitude[/tex]
[tex]\omega =angular\ frequency[/tex]
t=time
thus velocity at any Point x is given by
[tex]v=\omega \Sqrt{A^2-x^2}[/tex]
when speed is 50 % of its maximum value
[tex]v=0.5 A\omega [/tex]
[tex]0.5 A\omega =\omega \Sqrt{A^2-x^2}[/tex]
[tex]\frac{A^2}{4}=A^2-x^2[/tex]
[tex]x=\frac{\sqrt{3}}{2}A[/tex]
(b)When Potential Energy is 50 % of total energy
Total Energy [tex]=\frac{1}{2}kA^2[/tex]
potential energy at any x is given by [tex]U=\frac{1}{2}kx^2[/tex]
[tex]U=0.5\times \frac{1}{2}kA^2[/tex]
[tex]\frac{1}{2}kx^2=0.5\times \frac{1}{2}kA^2[/tex]
[tex]x=\frac{A}{\sqrt{2}}[/tex]
The answer is "0.95A and 0.548A", and the following are the calculation to the given points:
For point A:
In terms of displacement, write the expression for the oscillating mass's speed.
[tex]\to v=\omega \sqrt{A^2-x^2} [/tex]
Write the maximum speed of the oscillating object.
[tex]\to V_{max} = \omega \sqrt{A^2 -0^2}\\\\ [/tex]
[tex]= \omega \sqrt{A^2 -0}\\\\ = \omega \sqrt{A^2}\\\\=\omega A\\\\ [/tex]
If the speed is at [tex]30\%[/tex] of its maximum,
[tex]30\% \ v_{max}= \omega \sqrt{A^2 - x^2} \\\\ \frac{30}{100} \omega A= \omega \sqrt{A^2-x^2}\\\\ 0.09A^2 =A^2- x^2\\\\ x^2= A^2(1-0.09)\\\\ x^2= \sqrt{A^2\ (0.91)}\\\\ x= A\sqrt{(0.91)}\\\\ x=0.95\ \ A\\[/tex]
For point B:
The elastic potential energy of the spring is [tex]30\%[/tex] of the overall energy oscillating system, thus write the expression for it.
[tex]30\% ( \frac{1}{2}\ KA^2)= \frac{1}{2}Kx^2\\\\ 0.3\ A^2= x^2\\\\ x=0.548\ A\\\\[/tex]
Therefore, the final answer is "0.95A and 0.548A".
Learn more:
brainly.com/question/12905910
