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Found in: Page 1147

Fundamentals Of Physics

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 $4.39\mathrm{\mu s}$.

See the step by step solution

Step 1: Describe the expression for length of contraction

The expression of time dilation is given by,

${\mathbf{∆}}{{\mathbf{x}}}^{{\mathbf{\text{'}}}}{\mathbf{=}}{\mathbf{\gamma }}\left(∆x-v∆t\right)$ ……. (1)

Step 2: Find the speed parameter of

(a)

In the frame S, the bigger flash occurs at x1 position and the smaller flash occurs at x2 position .

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

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

$∆{\mathrm{x}}^{\text{'}}=0$

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

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

Substitute 1200 m for x1, 480 m for x2, and $5.00\mathrm{\mu s}$ for $∆\mathrm{t}$ in equation (2).

The expression to calculate speed parameter is given by,

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

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

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

Therefore, the speed parameter is 0.48.

Step 3: Determine whether S' moving in the positive or negative direction of the x axis

(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.

Step 4: Find the flash that occur first

(c)

The expression to calculate the time dilation is given by,

……. (4)

Substitute 1200 m for x1, 480 m for x2, $5.00\mathrm{\mu s}$ for $∆\mathrm{t}$, $1.44×{10}^{8}\mathrm{m}/\mathrm{s}$ for v, $3×{10}^{8}\mathrm{m}/\mathrm{s}$ for c, and 0.48 for in equation (4).

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

Therefore, the bigger flash occurs first.

Step 5: Find the time interval between the flashes

(d)

Substitute 1200 m for x1, 480 m for x2, $5.00\mathrm{\mu s}$ for $∆\mathrm{t}$, $1.44×{10}^{8}\mathrm{m}/\mathrm{s}$ for v, $3×{10}^{8}\mathrm{m}/\mathrm{s}$ for c, and 0.48 for $\mathrm{\beta }$ in equation (4).

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