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Standard XII

Physics

Relative Velocity in 2D

A train S 1 m...

Question

A train $S_{1}$ moving with a uniform velocity of $108 k m / h$ approaches another train $S_{2}$ standing on a platform. An observer $O$ moves with a uniform velocity of $36 k m / h$ towards $S_{2}$ as shown in figure. Both the trains are blowing whistles of same frequency $120 H z$ . When O is $600 m$ away from $S_{2}$ and distance between $S_{1}$ and $S_{2}$ is $800 m$ , the number of beats heard by O is . (Speed of the sound $= 330 m / s$ )

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Solution

Speed of sound $= 330 m / s$
Number of beats = frequency difference between the frequency observed by observer from $S_{1}$ and $S_{2}$ , $f_{b} = 120 [(\frac{330 + 10 cos 53^{o}}{330 - 30 cos 37^{o}}) - (\frac{330 + 10}{330})] = 120 [\frac{336}{306} - \frac{34}{33}] = 8.128 H z$

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Q. A train

S_{1}

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S_{2}

standing on a platform. An observer

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Q. A train

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approaches another train

S_{2}

standing on a platform. An observer

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as shown in figure. Both the trains are blowing whistles of same frequency

120 H z

. When O is

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away from

S_{2}

and distance between

S_{1}

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Q. One train is approaching an observer at rest and another train is receding from him with the same velocity 4 m/s. Both the trains blow whistles of same frequency of 243 Hz. The beat frequency in Hz as heard by the observer is (speed of sound in air is 320 m/s)

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Speed of sound =330m/s Number of beats = frequency difference between the frequency observed by observer from S1 and S2, fb=120[(330+10cos53o330−30cos37o)−(330+10330)]=120[336306−3433]=8.128Hz

Speed of sound $= 330 m / s$
Number of beats = frequency difference between the frequency observed by observer from $S_{1}$ and $S_{2}$ , $f_{b} = 120 [(\frac{330 + 10 cos 53^{o}}{330 - 30 cos 37^{o}}) - (\frac{330 + 10}{330})] = 120 [\frac{336}{306} - \frac{34}{33}] = 8.128 H z$