StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

25P

Expert-verifiedFound in: Page 1147

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**In Fig. 37-9, observer S detects two flashes of light. A big flash occurs at ${{\mathbf{x}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{1200}}{\mathbf{}}{\mathbf{m}}$ and, ${\mathbf{5}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{\mu s}}$ later, a small flash occurs at ${{\mathbf{x}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{480}}{\mathbf{}}{\mathbf{m}}$. As detected by observer S^{'}, the two flashes occur at a single coordinate x^{'}. (a) What is the speed parameter of S^{'}, and (b) is S^{'} moving in the positive or negative direction of the x axis? To S^{'}, (c) which flash occurs first and (d) what is the time interval between the flashes?**

(a) The speed parameter is 0.48.

(b) The S^{'} frame is moving in the negative x-direction.

(c) The bigger flash occurs first.

(d) The time interval is $4.39\mathrm{\mu s}$.

**The expression of time dilation is given by,**

** **

**${\mathbf{\u2206}}{{\mathbf{x}}}^{{\mathbf{\text{'}}}}{\mathbf{=}}{\mathbf{\gamma}}{\left(\u2206x-v\u2206t\right)}$ ……. (1)**

(a)

** **

In the frame S, the bigger flash occurs at x_{1} position and the smaller flash occurs at x_{2} position .

$\u2206\mathrm{x}={\mathrm{x}}_{1}-{\mathrm{x}}_{2}$

In the frame S^{'} both the flashes occur at the same time.

$\u2206{\mathrm{x}}^{\text{'}}=0$

Substitute $\u2206{\mathrm{x}}^{\text{'}}=0$ and $\u2206\mathrm{x}={\mathrm{x}}_{1}-{\mathrm{x}}_{2}$ in equation (1).

$0=\mathrm{\gamma}\left({\mathrm{x}}_{1}-{\mathrm{x}}_{2}-\mathrm{v}\u2206\mathrm{t}\right)\phantom{\rule{0ex}{0ex}}\mathrm{v}=\frac{{\mathrm{x}}_{1}-{\mathrm{x}}_{2}}{\u2206\mathrm{t}}$ …… (2)

Substitute 1200 m for x_{1}, 480 m for x_{2}, and $5.00\mathrm{\mu s}$ for $\u2206\mathrm{t}$ in equation (2).

$\begin{array}{rcl}\mathrm{v}& =& \frac{1200\mathrm{m}-480\mathrm{m}}{\left(500\mathrm{\mu s}\right)\left(\frac{{10}^{-6}\mathrm{s}}{1\mathrm{\mu s}}\right)}\\ & =& 1.44\times {10}^{8}\mathrm{m}/\mathrm{s}\end{array}$

The expression to calculate speed parameter is given by,

$\mathrm{\beta}=\frac{\mathrm{v}}{\mathrm{c}}$ ……. (3)

Substitute $1.44\times {10}^{8}\mathrm{m}/\mathrm{s}$ for v, and $3\times {10}^{8}\mathrm{m}/\mathrm{s}$ for c in equation (3).

$\begin{array}{rcl}\mathrm{\beta}& =& \frac{1.44\times {10}^{8}\mathrm{m}/\mathrm{s}}{3\times {10}^{8}\mathrm{m}/\mathrm{s}}\\ & =& 0.48\end{array}$

Therefore, the speed parameter is 0.48.

(b)

The speed is positive, so, the direction of S^{'} frame must be along the negative x-axis. Therefore, the S^{'} frame is moving in the negative x-direction.

(c)

The expression to calculate the time dilation is given by,

$\begin{array}{rcl}\u2206{\mathrm{t}}^{\text{'}}& =& \frac{\left(\u2206\mathrm{t}-\frac{\mathrm{v}\u2206\mathrm{x}}{{\mathrm{c}}^{2}}\right)}{\sqrt{1-{\mathrm{\beta}}^{2}}}\\ & =& \frac{\left(\u2206\mathrm{t}-\frac{\mathrm{v}\left({\mathrm{x}}_{1}-{\mathrm{x}}_{2}\right)}{{\mathrm{c}}^{2}}\right)}{\sqrt{1-{\mathrm{\beta}}^{2}}}\end{array}$ ……. (4)

Substitute 1200 m for x_{1}, 480 m for x_{2}, $5.00\mathrm{\mu s}$ for $\u2206\mathrm{t}$, $1.44\times {10}^{8}\mathrm{m}/\mathrm{s}$ for v, $3\times {10}^{8}\mathrm{m}/\mathrm{s}$ for c, and 0.48 for in equation (4).

$\begin{array}{rcl}\u2206{\mathrm{t}}^{\text{'}}& =& \frac{\left(5.00\mathrm{\mu s}\left(\frac{{10}^{-6}\mathrm{s}}{1\mathrm{\mu s}}\right)-\frac{\left(1.44\times {10}^{8}\mathrm{m}/\mathrm{s}\right)\left(1200\mathrm{m}-480\mathrm{m}\right)}{{\left(3\times {10}^{8}\mathrm{m}/\mathrm{s}\right)}^{2}}\right)}{\sqrt{1-{\left(0.48\right)}^{2}}}\\ & =& 4.386\times {10}^{-6}\mathrm{s}\left(\frac{1\mathrm{\mu s}}{{10}^{-6}\mathrm{s}}\right)\\ & =& 4.39\mathrm{\mu s}\end{array}$

The time is positive and hence, the bigger flash occurs first.

Therefore, the bigger flash occurs first.

(d)

Substitute 1200 m for x_{1}, 480 m for x_{2}, $5.00\mathrm{\mu s}$ for $\u2206\mathrm{t}$, $1.44\times {10}^{8}\mathrm{m}/\mathrm{s}$ for v, $3\times {10}^{8}\mathrm{m}/\mathrm{s}$ for c, and 0.48 for $\mathrm{\beta}$ in equation (4).

$\begin{array}{rcl}\u2206{\mathrm{t}}^{\text{'}}& =& \frac{\left(5.00\mathrm{\mu s}\left(\frac{{10}^{-6}\mathrm{s}}{1\mathrm{\mu s}}\right)-\frac{\left(1.44\times {10}^{8}\mathrm{m}/\mathrm{s}\right)\left(1200\mathrm{m}-480\mathrm{m}\right)}{{\left(3\times {10}^{8}\mathrm{m}/\mathrm{s}\right)}^{2}}\right)}{\sqrt{1-{\left(0.48\right)}^{2}}}\\ & =& 4.386\times {10}^{-6}\mathrm{s}\left(\frac{1\mathrm{\mu s}}{{10}^{-6}\mathrm{s}}\right)\\ & =& 4.39\mathrm{\mu s}\end{array}$

Therefore, the time interval is $4.39\mathrm{\mu s}$.

94% of StudySmarter users get better grades.

Sign up for free