# Mechanics Homework, Set 2

1. A 0.80 m -long metal rod is dropped vertically onto a marble floor. When the rod strikes the floor, it is determined electronically that the impact produces a 5 kHz-tone.What is the speed of sound in the rod? Express your answer using two significant figures.

2. The speed of sound in steel is about 4.50 k/ms A steel rail is struck with a hammer, and an observer0.420 km away has one ear to the rail. A) How much time will elapse from the time the sound is heard through the rail until the time it is heard through the air? Assume that the air temperature is 20 degrees and that no wind is blowing. B) How much time would elapse if the wind were blowing toward the observer at 34.0 km/h from where the rail was struck?

3. On hiking up a mountain that has several overhanging cliffs, a climber drops a stone at the first cliff to determine its height by measuring the time it takes to hear the stone hit the ground. A) On hiking up a mountain that has several overhanging cliffs, a climber drops a stone at the first cliff to determine its height by measuring the time it takes to hear the stone hit the ground. A) If the measured time is 4.30 for the stone dropping from the first cliff, and the air temperature is 20 C, how high is the cliff? B) If the height of a third cliff is three times that of the first one, what would be the measured time for a stone dropped from that cliff to reach the ground?

4. You are driving east at 25.0 m/s as you notice an ambulance traveling west toward you at 35.0 m/s . The sound you detect from the sirens has a frequency of 310 Hz. A)Determine the true frequency of the sirens. Assume normal room temperature.

5. An open organ pipe is 0.66 m long. A) If the speed of sound is 340 m/s , what is the pipe’s fundamental frequency? B) What are the frequencies of the first two overtones? Express your answers using two significant figures separated by a comma.

1. The rod will vibrate having at both ends maximum amplitude and at the center one node. Hence it will have only one half of the wavelength ($\lambda$)

$\lambda = 2*L = 2*0.8 =1.6 m$

$\lambda = wave speed * Time period =speed/ frequency$

$speed = \lambda*frequency = 2*L*frequency = 2*0.8*5000 = 8000 m/s$

2. A)

sound speed in air $V_{air} = 343 m/s$ at 20 degree Celsius

$time = -distance/V_{steel} +distance/V_{air} =-420/4500 +420/343 =1.13 seconds$

B)

wind speed $V_{wind} =34000/3600 =9.44 m/s$

because the sound is moving through the air $V_{sound} = V_{air}+V_{wind} =343+9.44 =352.44 m/s$

$time = distance/V_{sound} -distance/V_{steel} =420/352.44 -420/4500 =1.10 seconds$

3.

Let $H$ be the height of the cliff. The problem can be solved in two ways.

First way is to observe that the speed of sound is much greater than the speed of the rock and therefore all the measured time is approximately the time of falling.

$H= g*t^2/2 =9.81*4.3^2/2 =90.69 m$

Second way is to take into consideration also the time needed for the sound to travel back from the ground to the observer. let t1 be the time of falling and t2 the time of the travel of the sound from the ground to the observer

$t_1+t_2 =t$

$H=g*t_1^2/2$, $t_1 =\sqrt{2*H/g}$

$H=V_{sound}*t_2$, $t_2 = H/V_{sound}$, $V_{sound} =343 m/s$

$H/V_{sound} +\sqrt{2*H/g} =t$

$\sqrt{2*H/9.81} =4.3 -H/343$

$0.204*H = (4.3-0.0029*H)^2$

$0.204*H = 18.49 +8.41*10^-6*H^2 -2.5*10^-2*H$

$8.41*10^-6*H^2 -0.23*H +18.49 =0$

$H =80.85 m$

B)

$H_2 =3*H = 3*80.85 =242.57 m$

Total time

$t = \sqrt{2*H_2/g} + H_2/V_{sound} =\sqrt{2*242.57/9.81} +242.57/343 =7.74 seconds$

4.

The Doppler effect is shifting the initial frequency $F_0$ towards $F$ with the amount

$F = (V+V_r)/(V+V_s) *F_0$,

where V is the speed of the wave in the medium

$V_r$ is the speed of the receiver positive if the receiver is moving towards the source

$V_s$ is the speed of the source positive if the source is moving away from the receiver

Hence (assuming 20 degree Celsius $V = 343 m/s$)

$V_{receiver} = +25 m/s$, $V_{source} = -35 m/s$

$310 = (343 +25)/(343-35) *F_0$

$310 =1.194*F_0$

$F_0 =259.456 Hz$

5.

Let the pipe be open at one end and closed at the other end. Then the length of the pipe is $\lambda/4$ ($\lambda = wavelength$)

$L= \lambda/4$

$\lambda = L*4 =0.66*4 =2.64 m$

$\lambda = v*T = v/F$ (T is the period, F the frequency)

$F = v/\lambda = 340/2.64 =128.79 Hz$

First overtone

$F_1= 2*F = 2*128.79 =257.58 Hz$

Second overtone

$F_2 =3*F = 3*128.79 =386.36 Hz$

Let the pipe be open at both ends. Then the length of the pipe is $|lambda/2$ ($|lambda = wavelength$)

$L= \lambda/2$

$lambda = L*2 =0.66*2 = 1.32 m$

$\lambda = v*T = v/F$ (T is the period, F the frequency)

$F = v/\lambda = 340/1.32 = 257.58 Hz$

First overtone

$F_1= 2*F = 2*257.58 = 515.15 Hz$

Second overtone

$F_2 =3*F = 3*257.58 = 772.72 Hz$