(!) A locomotive engine approaches a railway station and whistles at a frequency of 400Hz .
A stationary observer on the platform observes a change of 40 Hz as the engine passes
across him . If the velocity of sound is 330m/s , the speed of the engine is ?
(A) 33 m/s
(B) 18 m/s
(C) 16.5 m/s
(D) 24 m/s
1i thnk its (a)
i got like this
for observer at rest and source movind towards observer f = f0(vv - vs) where vs is speed of source and v is speed of sound
now f = f0(vv - vs)
→ ff0 = vv - vs
→ f0f = 1 - vsv
→ vsv = f - fof
→ vs = (f - fof)v
= 40 x 330400 = 33 ms-1 as given that (f - fo = 40 Hz )
1Ans given is (c)
Thats y i posted the qs
So shud i take it for granted (from the pink abv) that ans given is wrong ??
1may be i need to reconsider the question as my answer is just double of the given isliye sayad kahin galti hogayi ho
[1]
62maya .. wrong pink i guess...
The question is about the change in frequency...
SO lets see that thing again... (1st assumption which i think the question setter did not give is tha tthe man is stading on the track.. If he stands on the platform, at a distance from the train, the change will be gradual and not immediate)
Now see the situation just before and after...
initially, v and vs are in the same direction....
After crossing, they are opposite...
Now, we have to take the difference..
Initially f = f0v(v-vs)
after crossing,
f = f0v(v+vs)
Now, the difference will be
f = f0[v(v-vs)-v(v+vs)]
f = f0[v.(2vs)(v2-vs2)]
Now try to figure out the answer?
1yes i made a mistake that when the train approaches then in the denomiantor it shud be V-Vs and when it leaves the observer it shud be V+Vs yes here i did a mistake[2]
1same thng i'm trying now and was going to post but nishant bhaiya posted.
I'm happy that even after making a mistake earlier i did correct in the 2nd try yippy [1][4]
u will make a quadratic in vs and solve i guess.
1@Nishant Sir thanks for the solution
Even i made the mistake of considering only one case