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3 years ago

5

You are standing at a subway platform when a subway passes, sounding its whistle. If you observe the whistle to be at 12750Hz wh

en it is approaching you and at 10750Hz after it passes,

1 answer:

adell [148]3 years ago

5 0

As per doppler's Effect of sound we can say when subway is approaching the platform we will have

$f_1 = f_o* \frac{v}{v- v_s}$

$12750 = f_o* \frac{340}{340 - v_s}$

Similarly we can find the frequency when subway is passing away with same speed from us

$f_2 = f_o* \frac{v}{v + v_s}$

$10750 * f_o* \frac{340}{340 + v_s}$

now we can find the ratio of two

$\frac{12750}{10750} = \frac{340 + v_s}{340 - v_s}$

$1.186*(340 - v_s) = 340 + v_s$

$403.26 - 1.186 v_s = 340 + v_s$

$63.26 = 2.186 v_s$

$v_s = \frac{63.26}{2.186}$

$v_s = 28.94 m/s$

So the speed of subway is 28.94 m/s

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