question_answer

When both the listener and source are moving towards each other, then which of the following is true regarding frequency and wavelength of wave observed by the observer?

A)  More frequency, less wavelength

B)  More frequency, more wavelength

C)  Less frequency, less wavelength

D)  More frequency, constant wavelength

Correct Answer: A

Solution :

Key Idea When the distance between the source and listener is decreasing, the apparent frequency increases. According to Dopplers effect, whenever there is a relative motion between a source of sound and listener, the apparent frequency of sound heard by the listener is different from the actual frequency of sound emitted by the source. Let S be source of sound and L the listener of sound. Let v be the actual frequency of sound emitted by the source and \[\lambda \] be the actual wavelength of the sound emitted. If v is velocity of sound in still air, then                 \[\lambda =\frac{v}{v}\] If velocity of listener is \[{{v}_{L}}\] and velocity of source is \[{{v}_{S}}\], then apparent frequency of sound waves heard by the listener is                 \[v=\frac{v-{{v}_{L}}}{v-{{v}_{S}}}\times v\] Here, both source and listener are approaching each other. Then \[{{v}_{S}}\] is positive-and \[{{v}_{L}}\] is negative. \[\therefore \] \[v=\frac{v-(-{{v}_{L}})}{v-{{v}_{S}}}v\]                 \[=\left( \frac{v+{{v}_{L}}}{v-{{v}_{S}}} \right)\,v\] ie,           \[v>v\] Also,      \[\lambda <\lambda \] So, listener, listens more frequency and observes less wavelength.