The problem is as follows:
A shinkansen which cruising with a constant rate across a railway in Tokyo emits a warning beep during $2\,s$. Find for how long (measured in seconds) a student standing right next to the railway will listen the sound. Assume the speed of the sound is $340\frac{m}{s}$.
The existing alternatives in my book are:
$\begin{array}{ll} 1.&\textrm{2 s}\\ 2.&\textrm{1.5 s}\\ 3.&\frac{17}{9}\textrm{s}\\ 4.&\frac{36}{17}\textrm{s}\\ 5.&\frac{17}{16}\textrm{s}\\ \end{array}$
I'm confused exactly how to solve this riddle?. How should I use the information of the speed of the sound here?. I'm assuming that the warning beep from the train will travel at constant rate and it will be "added" because of the reference frame?.
This part is confusing.
There isn't any information given with respect of the speed of the shinkansen. It only states that the beep is for two seconds. How exactly does this helps in the solution?.
I'm only able to find the distance covered by the beep in that amount of lenght which is:
$\frac{1}{340}\frac{m}{s}\times 2\,s= 680\,m$
But this doesn't give enough information to be used, other than is just known. Can someone help me here?.
I've been tempted to believe naively that the answer is just for $2\,s$ and this checks with the book as the correct answer.
But I don't know any mathematical justification. Does it exist any?. Is it related with doppler effect or what?.
It is difficult for me with agree with the answer given.
The simplest logic says that the time heard will be $T=2(1-v_c/v)$, where $v_c$ is the cruising speed.
The explanation is, assuming that the stating observer and the moving source (approaching the observer) are on the same straight line (see figure),
[the image taken from Physics by Resnick and Halliday, 1968, clearly illustrates the contraction of the wavefronts]
At $t_i=0$, the first wavefront starts when the source is at $S_1$. It reaches the observer at $t_1=d/v$, where $\overline{S_1O} = d$ and $v=340m/s$.
At $t_f=2$, the last wavefront emits when the source is at $S_2$. It reaches the observer after a time interval of $\Delta t=[d-v_c(t_f-t_i)]/v = t_1-2v_c/v$. The observer hears this wavefront at $t_2=2+\Delta t$.
[Note, $\overline{S_2O} = \overline{S_1O} - \overline{S_1S_2} = d - v_c(t_f-t_i)$]
[Also Note, the speed of each wavefront is $v$, since, once it leaves the source it travels through air at sound speed]
Therefore, the duration for which the observer hears the sound it $T=t_2 - t_1=2+t_1-2v_c/v-t_1=2(1-v_c/v)$.