Chapter 17 – Student Solutions Manual 5. Let tfbe the time for the stone to fall to the water and tsbe the time for the sound of the splash to travel from the water to the top of the well. Then, the total time elapsed from dropping the stone to hearing the splash is t= tf+ ts. If dis the depth of the well, then the kinematics of free fall gives 212fdgt=, or 2/.ftd=gThe sound travels at a constant speed vs, so d= vsts, or ts= d/vs. Thus the total time is 2//stdgd=+v. This equation is to be solved for d. Rewrite it as 2//sdgtdv=−and square both sides to obtain 2d/g= t2– 2(t/vs)d+ (1 + 2sv)d2. Now multiply by g2svand rearrange to get gd2– 2vs(gt+ vs)d+ g2svt2= 0. This is a quadratic equation for d. Its solutions are ()2222()44.2sssssvgtvvgtvg v tdg+±+−=22The physical solution must yield d= 0 for t= 0, so we take the solution with the negative sign in front of the square root. Once values are substituted the result d= 40.7 m is obtained. 7. If dis the distance from the location of the earthquake to the seismograph and vsis the speed of the S waves then the time for these waves to reach the seismograph is ts. = d/vs. Similarly, the time for P waves to reach the seismograph is tp= d/vp. The time delay is Δt= (d/vs) – (d/vp) = d(vp– vs)/vsvp, so 3(4.5 km/s)(8.0km/s)(3.0min)(60s /min)1.910 km.()8.0km/s4.5km/ssppsv vtdvvΔ===×−−We note that values for the speeds were substituted as given, in km/s, but that the value for the time delay was converted from minutes to seconds. 9. (a) Using λ= v/f, where vis the speed of sound in air and fis the frequency, we find 56343m/s7.6210m.4.5010 Hz−λ ==××
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