Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

In Fig. 37-9, observer S detects two flashes of light. A big flash occurs at $x_{1} = 1200 m$ and, $5.00 μs$ later, a small flash occurs at $x_{2} = 480 m$ . 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.

(d) The time interval is $4.39 μs$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Describe the expression for length of contraction

The expression of time dilation is given by,

$∆ x^{'} = γ (∆ x - v ∆ t)$ ……. (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 .

$∆ x = x_{1} - x_{2}$

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

$∆ x^{'} = 0$

Substitute $∆ x^{'} = 0$ and $∆ x = x_{1} - x_{2}$ in equation (1).

$0 = γ (x_{1} - x_{2} - v ∆ t) v = \frac{x_{1} - x_{2}}{∆ t}$ …… (2)

Substitute 1200 m for x₁, 480 m for x₂, and $5.00 μs$ for $∆ t$ in equation (2).

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

The expression to calculate speed parameter is given by,

$β = \frac{v}{c}$ ……. (3)

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

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

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,

$\begin{array}{rcl} ∆ t^{'} & = & \frac{(∆ t - \frac{v ∆ x}{c^{2}})}{\sqrt{1 - β^{2}}} \\ = & \frac{(∆ t - \frac{v (x_{1} - x_{2})}{c^{2}})}{\sqrt{1 - β^{2}}} \end{array}$ ……. (4)

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

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

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, $5.00 μs$ for $∆ t$ , $1.44 \times 10^{8} m / s$ for v, $3 \times 10^{8} m / s$ for c, and 0.48 for $β$ in equation (4).

Therefore, the time interval is $4.39 μs$ .

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Describe the expression for length of contraction

Step 2: Find the speed parameter of

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

Step 4: Find the flash that occur first

Step 5: Find the time interval between the flashes

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Describe the expression for length of contraction

Step 2: Find the speed parameter of

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

Step 4: Find the flash that occur first

Step 5: Find the time interval between the flashes

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