contestada

According to observer O, a blue flash occurs at xb = 10.4 m when tb= 0.124 μs, and a red flash occurs at xr= 23.6 m when tr= 0.138 μs. According to observer O’, who is in motion relative to O at velocity u, the two flashes appear to be simultaneous. Find the velocity u.

Respuesta :

Answer:

The velocity is  [tex]u=0.3m/s[/tex]

Explanation:

From the question we can deduce that both flashes occurred simultaneous to the observer

    So the first derivative of the time difference   [tex]\Delta t' =0[/tex]

Generally the Lorentz transformation i mathematically denoted as

               [tex]\Delta t' = \frac{ \Delta t -\frac{v \Delta x}{c^2} }{\sqrt{1-\frac{v^2}{c^2} } }[/tex]

=>           [tex]0 = \frac{ \Delta t -\frac{v \Delta x}{c^2} }{\sqrt{1-\frac{v^2}{c^2} } }[/tex]

=>           [tex]\Delta t -\frac{u \Delta x }{c^2} = 0[/tex]

Now making u the subject

             [tex]u = \frac{c^2 \Delta t }{\Delta x}[/tex]

                 [tex]= \frac{t_b - t_r}{x_b - x_r} c^2[/tex]

substituting [tex]0.24 \mu s = 0.24*10^{-6}s[/tex] for [tex]t_b[/tex] , [tex]0.138 \mu s = 0.138*10^{-6}s[/tex] for [tex]t_r[/tex] , [tex]10.4 m[/tex] for [tex]x_b[/tex] , [tex]23.6 m[/tex] for [tex]x_r[/tex] and  [tex]3.0*10^8 m/s[/tex] for c

       

              [tex]u = \frac{0.124*10^ {-6} -0.138 *10^{-6}}{10 - 23.6}[/tex]

                [tex]u=0.3m/s[/tex]

Answer:

u = 95.46 * 10⁶ m/s

Explanation:

For the first observer O,

Time of the blue flash, [tex]t_{b} = 0.124 \mu s[/tex]

Time of the red flash, [tex]t_{r} = 0.138 \mu s[/tex]

Distance covered before the blue flash, [tex]x_{b} = 10.4 m[/tex]

Distance covered before the red flash,  [tex]x_{r} = 23.6 m[/tex]

Difference in time observed by the first observer O,  Δt = [tex]t_{r} - t_{b}[/tex]

Δt = 0.138 - 0.124 = 0.014  μs

Difference in the distance observed by the observer O, Δx = [tex]x_{r} - x_{t}[/tex]

Δx = 23.6 m - 10.4 m

Δx = 13.2 m

Since for the observer O', the two flashes are simultaneous, there will be no time difference between the two flashes. i.e. Δt' = 0

Using the lorent'z transformation equation:

[tex]\triangle t' = \frac{\triangle t - \frac{u \triangle x}{c^{2} } }{\sqrt{1 - \frac{u^{2} }{c^{2} } } }[/tex]...................(1)

But Δt' = 0,

equation (1) becomes

[tex]0 = \triangle t - \frac{u \triangle x}{c^{2} } \\[/tex]

[tex]0 = (0.014 * 10^{-6} ) - \frac{u * 13.2}{(3*10^{8}) ^{2} }[/tex]

[tex](0.014 * 10^{-6} ) = \frac{u * 13.2}{(3*10^{8}) ^{2} } \\u = 95454545.46 m/s[/tex]

u = 95.46 * 10⁶ m/s