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A small source of sound oscillates in simple harmonic motion with an amplitude of 17 cm. A detector is placed along the line of motion of the source. The source emits a sound of frequency 800 Hz which travels at a speed of `340 ms^-1` If the width of the frequency band detected by the detector is 8 Hz, find the time period of the source. |
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Answer» Correct Answer - C Given that `r=0.17m, f=800Hz, u=340m/s` Frequency band `=f_1+f_2=8 Hz` Where `f_1 and f_2` correspond to the maximum and minimum apprent frequencies.(Both will be the mean positioin, because the velocity is maximum). `Now, f_1=(34/(340+v_s))f` `f_2=(340/((340-v_s))f` `:.f_1-f_2=8` `=340[1/(340-v_s)-1/(340+v_s)]` =8 `rarr (2vs)/((340s^2-v_s^2)=8/(340xx800)` `rarr 340^2-vs^2=68000vs` solving for `v_s` we get `vs=1.695m/s` for SHM `v_s=rw` `rarr omega=(1.695/0.17)=10` So, T=(2pi)/w=pi/5=0.63 sec |
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