A train S1, moving with a uniform velocity of 108 km/h, approaches another train S2 standing on a platform. An observer O moves with a uniform velocity of 36 km/h towards S2, as shown in figure. Both the trains are blowing whistles of same frequency 120 Hz. When O is 600 m away from S2 and distance between S1 and S2 is 800 m, the number of beats heard by O is ............ . [Speed of the sound = 330 m/s ............ .]

Waves PYQ — JEE Advanced 2019 · Shift 1 · Q54 with Solution

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Correct answer: 8.13

Convert all speeds into m/s

Speed of train $S_1$ : $108\,\text{km/h} = 30\,\text{m/s}$
Speed of observer $O$ : $36\,\text{km/h} = 10\,\text{m/s}$
Speed of sound: $v = 330\,\text{m/s}$
Frequency of both whistles: $f = 120\,\text{Hz}$

Interpret the geometry and motions

Train $S_2$ is standing on the platform.
Observer $O$ moves towards $S_2$ with speed $10\,\text{m/s}$ .
Train $S_1$ moves towards $S_2$ with speed $30\,\text{m/s}$ .

Given at the instant considered:

Distance of $O$ from $S_2$ = $600\,\text{m}$
Distance between $S_1$ and $S_2$ = $800\,\text{m}$

So, if all are on the same line, then distance between $S_1$ and $O$ is $800 - 600 = 200\,\text{m}$ Thus $S_1$ is behind $O$ , and both are moving toward $S_2$ .

Find apparent frequency heard from stationary train $S_2$

For a stationary source and observer moving towards the source: $f_2 = f\left(\frac{v+v_O}{v}\right)$ Here $v_O=10\,\text{m/s}$ , so

=120\left(\frac{340}{330}\right) =123.636\,\text{Hz}$$ --- 4. **Find apparent frequency heard from moving train $S_1$** Train $S_1$ is moving towards observer $O$, so wavelength in front of source is shortened. Frequency heard by moving observer is: $$f_1 = f\left(\frac{v-v_O'}{v-v_S}\right)$$ A safer way is to use sign physically: - Observer is moving **away from $S_1$** (since both move right, but observer lies in front of $S_1$), so effective numerator is $v-v_O = 330-10=320$. - Source $S_1$ is moving **towards observer**, so denominator is $v-v_S = 330-30=300$. Hence $$f_1 = 120\left(\frac{320}{300}\right)=128\,\text{Hz}$$ --- 5. **Beat frequency at that instant** $$n = |f_1-f_2| = |128-123.636| = 4.364\,\text{Hz}$$ This is the beat frequency at the given instant. --- 6. **Time for observer to reach $S_2$** Observer is $600\,\text{m}$ away from $S_2$ and moves toward it with speed $10\,\text{m/s}$. So time taken is $$t=\frac{600}{10}=60\,\text{s}$$ --- 7. **Number of beats heard in this interval** Assuming this beat frequency remains constant over this interval, $$N = n\,t = 4.364\times 60 = 261.84$$ This is not close to the stored answer, so let us check whether the question instead asks only for the **number of beats per second** (which is standard usage in many problems). Then the answer would be $$n=4.36$$ which still does not match $8.13$. --- 8. **Re-check using alternative interpretation of observer direction** If observer moves towards $S_2$, and $S_1$ is on the other side of $O$, then for sound from $S_1$, observer is moving **towards** $S_1$. Then $$f_1 = 120\left(\frac{330+10}{330-30}\right) =120\left(\frac{340}{300}\right)=136\,\text{Hz}$$ And from $S_2$: $$f_2=123.636\,\text{Hz}$$ So beat frequency becomes $$n=136-123.636=12.364\,\text{Hz}$$ Still not $8.13$. If instead one source is receding and the other approaching in another sign convention, no consistent geometry yields $8.13$ from the given data. --- 9. **Conclusion** From the standard Doppler interpretation of the stated geometry, the beat frequency should be $$\boxed{4.36\,\text{Hz}}$$ if the observer is between $S_1$ and $S_2$ and moving toward $S_2$. The distances given do not affect the instantaneous beat frequency, except to determine relative placement. Therefore, the stored answer $8.13$ does not agree with the physics of the problem as stated.

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View written solutionFree

Correct answer: 8.13

Convert all speeds into m/s

Speed of train $S_1$ : $108\,\text{km/h} = 30\,\text{m/s}$
Speed of observer $O$ : $36\,\text{km/h} = 10\,\text{m/s}$
Speed of sound: $v = 330\,\text{m/s}$
Frequency of both whistles: $f = 120\,\text{Hz}$

Interpret the geometry and motions

Train $S_2$ is standing on the platform.
Observer $O$ moves towards $S_2$ with speed $10\,\text{m/s}$ .
Train $S_1$ moves towards $S_2$ with speed $30\,\text{m/s}$ .

Given at the instant considered:

Distance of $O$ from $S_2$ = $600\,\text{m}$
Distance between $S_1$ and $S_2$ = $800\,\text{m}$

So, if all are on the same line, then distance between $S_1$ and $O$ is $800 - 600 = 200\,\text{m}$ Thus $S_1$ is behind $O$ , and both are moving toward $S_2$ .

Find apparent frequency heard from stationary train $S_2$

For a stationary source and observer moving towards the source: $f_2 = f\left(\frac{v+v_O}{v}\right)$ Here $v_O=10\,\text{m/s}$ , so

=120\left(\frac{340}{330}\right) =123.636\,\text{Hz}$$ --- 4. **Find apparent frequency heard from moving train $S_1$** Train $S_1$ is moving towards observer $O$, so wavelength in front of source is shortened. Frequency heard by moving observer is: $$f_1 = f\left(\frac{v-v_O'}{v-v_S}\right)$$ A safer way is to use sign physically: - Observer is moving **away from $S_1$** (since both move right, but observer lies in front of $S_1$), so effective numerator is $v-v_O = 330-10=320$. - Source $S_1$ is moving **towards observer**, so denominator is $v-v_S = 330-30=300$. Hence $$f_1 = 120\left(\frac{320}{300}\right)=128\,\text{Hz}$$ --- 5. **Beat frequency at that instant** $$n = |f_1-f_2| = |128-123.636| = 4.364\,\text{Hz}$$ This is the beat frequency at the given instant. --- 6. **Time for observer to reach $S_2$** Observer is $600\,\text{m}$ away from $S_2$ and moves toward it with speed $10\,\text{m/s}$. So time taken is $$t=\frac{600}{10}=60\,\text{s}$$ --- 7. **Number of beats heard in this interval** Assuming this beat frequency remains constant over this interval, $$N = n\,t = 4.364\times 60 = 261.84$$ This is not close to the stored answer, so let us check whether the question instead asks only for the **number of beats per second** (which is standard usage in many problems). Then the answer would be $$n=4.36$$ which still does not match $8.13$. --- 8. **Re-check using alternative interpretation of observer direction** If observer moves towards $S_2$, and $S_1$ is on the other side of $O$, then for sound from $S_1$, observer is moving **towards** $S_1$. Then $$f_1 = 120\left(\frac{330+10}{330-30}\right) =120\left(\frac{340}{300}\right)=136\,\text{Hz}$$ And from $S_2$: $$f_2=123.636\,\text{Hz}$$ So beat frequency becomes $$n=136-123.636=12.364\,\text{Hz}$$ Still not $8.13$. If instead one source is receding and the other approaching in another sign convention, no consistent geometry yields $8.13$ from the given data. --- 9. **Conclusion** From the standard Doppler interpretation of the stated geometry, the beat frequency should be $$\boxed{4.36\,\text{Hz}}$$ if the observer is between $S_1$ and $S_2$ and moving toward $S_2$. The distances given do not affect the instantaneous beat frequency, except to determine relative placement. Therefore, the stored answer $8.13$ does not agree with the physics of the problem as stated.

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