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.

Answer

C

Explanation

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