Respuesta :
Answer:
(a) Peak=2 cm, velocity= 3 cm/s (to the right)
(b) Peak= 2 cm, velocity= -3 cm/s (to the left)
Explanation:
Pulse Movement
The function of a moving wave is given by
[tex]\displaystyle y(x,t)=\frac{2}{(x-3t)^2+1}[/tex]
where x,y are measured in cm and t in seconds.
Please check the graph shown in the image provided below. It shows the waveform at two different times, t=0 sec and t=1 sec. The peak value is displaced by 3 cm when t varies by 1 second. It shows a velocity of 3 m/s.
(a) For a given time, say t=0 the expression for the pulse is
[tex]\displaystyle y(x,0)=\frac{2}{x^2+1}[/tex]
The maximum value or the pulse amplitude occurs when the denominator has its minimum value, that is when x=0
[tex]\displaystyle A=y(0,0)=\frac{2}{0^2+1}=2[/tex]
At t=1 second, the function is
[tex]\displaystyle y(x,1)=\frac{2}{(x-3)^2+1}[/tex]
Again, the maximum value of the wave occurs at the minimum value of the denominator, or when
[tex]x-3=0 \rightarrow x=3[/tex]
Note the peak has moved 3 cm to the right when t increased by 1 second, this gives us a horizontal velocity of 3 cm/s.
We can corroborate for the general case knowing the peak value moves to the right at the point where
[tex]x-3t=0 \rightarrow x=3t[/tex]
Taking the derivative with respect to t gives us the horizontal velocity:
[tex]x'=3\ cm/s[/tex]
At t=2 seconds
[tex]\displaystyle y(x,2)=\frac{2}{(x-6)^2+1}[/tex]
The amplitude and velocity are the same as determined before. The graph shown in the figure attached shows the pulse waves at t=0 and t=1
b) If the function was
[tex]\displaystyle y(x,t)=\frac{2}{(x+3t)^2+1}[/tex]
Then when t increases, the peak value moves to the left on the x-axis. The velocity would be
[tex]x'=-3\ cm/s[/tex]
It means the wave is traveling to the left instead of to the right
