Answer:
(a). 19.0 m/s
(b). 9.5m/s
Explanation:
Assuming speed of sound is 343m/s.
(a).
As the train approaches, from the the Doppler equation we have
[tex]f_{obs} = \dfrac{vf_{source}}{v-v_{source}}[/tex]
[tex]350Hz = \dfrac{343f_{source}}{343-v_{source}}[/tex]
solving for [tex]f_{source}[/tex] we get:
[tex]f_{source} = \dfrac{350(343-v_{source})}{343}[/tex].
And as the train cuts its speed in half, the equation gives
[tex]340Hz = \dfrac{343f_{source}}{343-\dfrac{v_{source}}{2} }[/tex]
substituting the value of [tex]f_{source}[/tex] we get:
[tex]340Hz = \dfrac{343\dfrac{350(343-v_{source})}{343}}{343-\dfrac{v_{source}}{2} }[/tex]
[tex]340Hz = \dfrac{350(343-v_{source})}{343-\dfrac{v_{source}}{2} }[/tex]
[tex]340Hz = \dfrac{700(343-v_{source})}{686-v_{source}}[/tex]
[tex]340 (686-v_{source}) = 700(343-v_{source})[/tex]
[tex]\boxed{v_{source} = 19.0m/s}[/tex]
or 68.6 km per hour, which is the speed of the train before slowing down.
(b).
The speed of the train after slowing down is half the previous speed; therefore,
[tex]v_{after} = \dfrac{v_{source}}{2}[/tex]
[tex]\boxed{v_{after} =9.5m/s}[/tex]
or 34.3 km per hour.
Given:
Train whistle,
Sound of whistle,
Speed of sound,
Now,
→ [tex]340 = F\times \frac{343}{343-v}[/tex]...(equation 1)
→ [tex]330 = F\times \frac{343}{343-\frac{v}{2} }[/tex]...(equation 2)
After dividing "equation 1" by "equation 2", we get
→ [tex]\frac{340}{330}= \frac{343-\frac{v}{2} }{343-v}[/tex]
By applying cross-multiplication, we get
→ [tex]1.0303030(343-v) = 343-\frac{v}{2}[/tex]
→ [tex]353.3939394-1.03030 v = 343-0.5 v[/tex]
[tex]10.3939394=0.5303 v[/tex]
[tex]v = \frac{10.3939394}{0.5303}[/tex]
[tex]= 19.600112 \ m/s[/tex] (Speed before train slowing down)
hence,
Speed after train slowing down will be:
= [tex]\frac{v}{2}[/tex]
= [tex]\frac{19.600112}{2}[/tex]
= [tex]9.800056 \ m/s[/tex]
Thus the responses above are correct.
Learn more about speed here:
https://brainly.com/question/14696414
