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FrozenT [24]
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,
Physics
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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