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A-train-travelling-at-30m-s-approaching-a-stationary-observer-emits-a-tone-at-a-frequency-of-480Hz-If-the-speed-of-sound-is-330m-s-determine-the-frequency-of-the-note-the-observer-hears-




Question Number 35881 by NECx last updated on 25/May/18
A train travelling at 30m/s ,  approaching a stationary observer  emits a tone at a frequency of 480Hz.  If the speed of sound is 330m/s,  determine the frequency of the  note the observer hears.
$${A}\:{train}\:{travelling}\:{at}\:\mathrm{30}{m}/{s}\:, \\ $$$${approaching}\:{a}\:{stationary}\:{observer} \\ $$$${emits}\:{a}\:{tone}\:{at}\:{a}\:{frequency}\:{of}\:\mathrm{480}{Hz}. \\ $$$${If}\:{the}\:{speed}\:{of}\:{sound}\:{is}\:\mathrm{330}{m}/{s}, \\ $$$${determine}\:{the}\:{frequency}\:{of}\:{the} \\ $$$${note}\:{the}\:{observer}\:{hears}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 25/May/18
Answered by tanmay.chaudhury50@gmail.com last updated on 25/May/18
f^′ =f(v/(v−v_s ))=480×((330)/(330−30))  f′=((480×330)/(300))  =48×11=528
$${f}^{'} ={f}\frac{{v}}{{v}−{v}_{{s}} }=\mathrm{480}×\frac{\mathrm{330}}{\mathrm{330}−\mathrm{30}} \\ $$$${f}'=\frac{\mathrm{480}×\mathrm{330}}{\mathrm{300}} \\ $$$$=\mathrm{48}×\mathrm{11}=\mathrm{528} \\ $$
Commented by NECx last updated on 28/May/18
Thank you Tanmay sir
$${Thank}\:{you}\:{Tanmay}\:{sir} \\ $$

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