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Cerrena [4.2K]
3 years ago
6

A 30-m-long rocket train car is traveling from Los Angeles to New York at 0.5c when a light at the center of the car flashes. Wh

en the light reaches the front of the car. it immediately rings a bell. Light reaching the back of the car immediately sounds a siren.
Part A
Are the bell and siren simultaneous events for a passenger seated in the car?
If not, which occurs first?
Yes, the bell and siren are simultaneous events. No, the siren sounds before the bell rings.
No, the bell rings before the siren sounds.
Part B
Are the bell and siren simultaneous events for a bicyclist waiting to cross the tracks?
If not, which occurs first?
Yes, the bell and siren are simultaneous events.
No, the siren sounds before the bell rings. No, the bell rings before the siren sounds.
Physics
1 answer:
Novosadov [1.4K]3 years ago
8 0

Given that,

Distance =30 m

speed = 0.5c

(A). We need to find the bell and siren simultaneous events for a passenger seated in the car

According to given data

The distance travelled by the light to reach either side of the rocket  train car is same.

So, The two events are simultaneous and the bell and siren are the simultaneous events for a passenger seated in the car.

(B). We need to calculate time interval between the events

Using formula of time dilation

\Delta t=\dfrac{\Delta t'}{\sqrt{1-\dfrac{v^2}{c^2}}}.....(I)

Where, \delta t' = proper time

\delta t = time interval between the events

The time interval between the events measured in a reference frame

The proper time in this case is

\Delta t'=\Delta t_{1}-\dfrac{v\Delta x}{c^2}

For the second interval,

Put the value of \Delta t' in the equation (I)

\Delta t_{2}=\dfrac{\Delta t_{1}-\dfrac{v\Delta x}{c^2}}{\sqrt{1-\dfrac{v^2}{c^2}}}

Put the value in the equation

\Delta t_{2} = \dfrac{0-\dfrac{0.5c\times30}{c^2}}{\sqrt{1-\dfrac{0.5^2c^2}{c^2}}}

\Delta t_{2}=\dfrac{-15}{3\times10^{8}\sqrt{1-0.25}}

\Delta t_{2}=-5.77\times10^{8}\ s

Negative sign shows the siren rings before the bell ring.

Hence, (A). Yes, the bell and siren are simultaneous events.

(B). The siren sounds before the bell rings.

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Answer:

Approximately 30.0\; \rm Hz.

Explanation:

Look up the speed of sounds in the air at 25\; ^\circ\rm C: v \approx 346\; \rm m \cdot s^{-1}.

Let the frequency of this tune be f\; \rm Hz. The wavelength of the tune would be \displaystyle \lambda = \frac{v}{f} \approx \frac{346}{f}.

The distance between the first speaker and the listener is 12.608\; \rm m. How many wavelengths can fit into that distance?

\displaystyle \frac{12.608}{\lambda} \approx \frac{12.608}{346 / f} = \frac{12.608}{346} \cdot f.

Similarly, the distance between the second speaker and the listener is 18.368\; \rm m. Number of wavelengths in that distance:

\displaystyle \frac{18.368}{\lambda} \approx \frac{18.368}{346 / f} = \frac{18.368}{346} \cdot f.

Difference between these two numbers:

\begin{aligned} &\frac{18.368}{346} \cdot f - \frac{12.608}{346} \cdot f = \frac{5.76}{346}\cdot f\end{aligned}.

For destructive interference to occur, that difference should be equal to \displaystyle \frac{1}{2}, \displaystyle 1 + \frac{1}{2}= \frac{3}{2}, \cdots, or  \displaystyle \left(k+ \frac{1}{2}\right) in general (k can be any non-negative whole number.)

Let \begin{aligned} \frac{5.76}{346}\cdot f = k + \frac{1}{2}\end{aligned}, and solve for f.

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The next step is to find the values of k that ensure 20 \le f \le 20\times 10^{3} (frequency is between 20\; \rm Hz and 20\, \rm kHz.) It turns out that k = 0 (the smallest k value possible) would be sufficient. In that case, the frequency is approximately 30.0\; \rm Hz. Using a larger k would only increase the frequency.

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Explanation:

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