2008 Semifinal Exam 8 Question A4 A tape recorder playing a single tone of frequency fo is dropped from rest at a height h.You stand directly underneath the tape recorder and measure the frequency observed as a function of time.Here t=0s is the time at which the tape recorder was dropped. t (s)f(Hz) 2.0 581 4.0 619 6.0 665 8.0 723 10.0 801 The acceleration due to gravity is g=9.80 m/s2 and the speed of sound in air is va =340 m/s.Ignore air resistance.You might need to use the Doppler shift formula for co-linear motion of sources and observers in still air, f=f%土 Va士vs where fo is the emitted frequency as determined by the source,f is the frequency as detected by the observer, and va,vs,and vo are the speed of sounds in air,the speed of the source,and the speed of the observer.The positive and negative signs are dependent upon the relative directions of the source and the observer. a.Determine the frequency measured on the ground at time t,in terms of fo,g,h,and va. b.Verify graphically that your result is consistent with the provided data. c.What (numerically)is the frequency played by the tape recorder? d.From what height h was the tape recorder dropped? Solution The position of the tape recorder above the ground at a time t is given by and the speed of the tape recorder is given by Us =-gt The observer "hears"the sound emitted from the tape recorder a time 6t earlier,since it takes time for the sound to travel to the listener.In this case. y=Vadt So at time t the listener is hearing the tape recorder when it had emitted at time t'=t-ot,or t=t-上+9e2 Va 2va Solve this for t',first by rearranging, y-w+(t-)=0 and the by applying the quadratic formula t=a±Va2+2gh-2gat 9 Copyright C2008 American Association of Physics TeachersSolutions 2008 Semifinal Exam 8 Question A4 A tape recorder playing a single tone of frequency f0 is dropped from rest at a height h. You stand directly underneath the tape recorder and measure the frequency observed as a function of time. Here t = 0s is the time at which the tape recorder was dropped. t (s) f (Hz) 2.0 581 4.0 619 6.0 665 8.0 723 10.0 801 The acceleration due to gravity is g = 9 .80 m / s 2 and the speed of sound in air is v a = 340 m /s. Ignore air resistance. You might need to use the Doppler shift formula for co-linear motion of sources and observers in still air, f = f0 v a ± v o v a ± v s where f0 is the emitted frequency as determined by the source, f is the frequency as detected by the observer, and v a , v s, and v o are the speed of sounds in air, the speed of the source, and the speed of the observer. The positive and negative signs are dependent upon the relative directions of the source and the observer. a. Determine the frequency measured on the ground at time t, in terms of f0 , g , h, and v a . b. Verify graphically that your result is consistent with the provided data. c. What (numerically) is the frequency played by the tape recorder? d. From what height h was the tape recorder dropped? Solution The position of the tape recorder above the ground at a time t is given by y = h − 12 gt 2 and the speed of the tape recorder is given by v s = −gt The observer “hears” the sound emitted from the tape recorder a time δt earlier, since it takes time for the sound to travel to the listener. In this case, y = v aδt So at time t the listener is hearing the tape recorder when it had emitted at time t ′ = t − δt, or t ′ = t − hva + g 2 v a ( t ′ ) 2 Solve this for t ′ , first by rearranging, g2 ( t ′ ) 2 − v a t ′ + ( v a t − h) = 0 and the by applying the quadratic formula t′ = v a ± p v a 2 + 2gh − 2gv a t g . Copyright c 2008 American Association of Physics Teachers