Physics Test (2010)

1) Two piano strings are supposed to be vibrating at 132 , but a piano tuner hears three beats every 2.9 when they are played together. A) If one is vibrating at 132 , what must be the frequency of the other (is there only one answer)? B) By how much (in percent) must the tension be increased or decreased to bring them in tune?


The frequency of beats is $f_b = 3 beats/2.9 seconds = 1.034 Hz$

There are two answers for the frequency of the second piano.

$F_2 = F_1+ f_b = 132+1.034 =133.034 Hz$

$F_2 = F_1 -f_b =132 -1.034 =130.966 Hz$


The velocity of the standing wave on a string having the mass per unit length $mu$ and the tension $T$ is

$v = \sqrt{T/\mu}$,

wavelength $\lambda = v*F$

and since the strings have the same length lambda is the same

$\sqrt {T_1}* F_1 = \sqrt{T_0}*F_0$

$\sqrt{T_1/T_0} = F_0/F_1$

$T_1/T_0 = (F_0/F_1)^2 =(132/133.034)^2 =0.9845$

$T_1 =98.45%*T_0$

$T_0-T_1 =1.548 % T_0$

$T_2/T_0 =(F_0/F_2)^2 =(132/130.966)^2 =1.0159$

$T_0-T_2 = 1 – 1.0159 =-0.0159 =-1.59% T$

2) A 70 person jumps from a window to a fire net 18 below, which stretches the net 1.0 . Assume that the net behaves like a simple spring. Calculate how much it would stretch if the same person were lying in it. How much would it stretch if the person jumped from 38 ?


The person speed when it touches the net is $V^2 =2gh$

$v = \sqrt{2*9.81*18} =18.79 m/s$

the deceleration during contact with net is

$V^2 =-2*a*d$

$-2ad =2gh$

$a =gh/d =9.81*18/1 = 176.58 m/s^2$

Force on net is $F = m*a = 70*176.58 =12360.6 N$

-time of contact is $t = v/a = 18.79/176.58 =0.106 sec$

-elastic constant of net is $K = F/d = 12360.6 N/m$

When the person lies on net the force is $F_1 =m*g =70*9.81 =686.7 N$

and the stretch of the net is $d_1 = F_1/K =686.7/12360.6 =0.0556 m =5.56 cm$


The speed at the contact with net is $V= \sqrt{2*g*h} =\sqrt{2*9.81*38}=27.3 m/s$

Kinetic energy is $E_c = m*v^2/2 =70*27.3^2/2 =26094.6 J$

this energy need to be equal to the potential energy of the net stretched

$E_p =k*x^2/2$

$x^2 =2*E_c/K = 2*26094.6/12360.6 =4.22 m^2$

$x =2.055 m$ the distance the net stretches

3) A violin string vibrates at 287 when unfingered. At what frequency will it vibrate if it is fingered one-third of the way down from the end? (That is, only two-thirds of the string vibrates as a standing wave.)


The wavelength is $L =v*T = v/F$ where $v$ is the speed of wave, $F$ the frequency, $T$ the period

$(2/3)* L= v/F_1$

by dividing one gets $3/2 = F_1/F$

$F_1 = 3/2*F =3/2*287 = 430.5 Hz$

4) Two violin strings are tuned to the same frequency, 294 . The tension in one string is then decreased by 1.5. What will be the beat frequency heard when the two strings are played together? [Hint: Use the equation ]


The velocity of the standing wave on a string having the mass per unit length $\mu$ and the tension $T$ is

$v = \sqrt{T/\mu}$,

wavelength $\lambda = v*F$

and since the strings have the same length $\lambda$ is the same

$\sqrt (T_1)* F_1 = \sqrt{T_2}*F_2$

$\sqrt(T_1/T_2) = F_2/F_1$

$F_2/F_1 = \sqrt{(100-1.5)/(100)} =\sqrt{98.5/100} =0.99247$

$F_2 =294*0.99247 =291.787 Hz$

F beats is $F = F_1-F_2 =294-291.787 =2.214 Hz$

5) If one vibration has 6.4 times the energy of a second, but their frequencies and masses are the same, what is the ratio of their amplitudes?


In general the energy $E$ is related to the amplitude of vibration by the relation

$E = 1/2*k*A^2*sin^2(\omega*t)$

$\omega$ is the same, $k$ is the same so that

$E_1/E_2 =(A_1/A_2)^2$

$A_1/A_2 = \sqrt(E_1/E_2) = \sqrt(6.4) =2.53$

$A_1 =2.53*A_2$

6) A stereo amplifier is rated at 130 output at 1000 . The power output drops by 11 at 15 . What is the power output in watts at 15 ?


The dB definition is

$dB =10*log(P/P_0)$ where $P_0$ is the reference power and log in in base 10

$-11 =10*log(P/P_0)$

$P/P_0 = 10^{-1.1} = 0.0794$

$P = 0.0794*130 =10.326 W$

In other books the dB definition is

$dB = 20*log(P/P_0)$

$-11 = 20*log(P/P_0)$

$P = P_0*10^(-0.55)$

$P =36.64 W$

7) A sound wave is traveling in warm air when it hits a layer of cold, dense air. If the sound wave hits the cold air interface at an angle of 20, what is the angle of refraction? Assume that the cold air temperature is -15 and the warm air temperature is +20. 


The speed of sound as a function of temperature can be approximated by, $v=331.3+0..606*t$  where is in Celsius degrees

$v(20 Celsius) =331+0.6*20 =343 m/s$

$v(-15 Celsius) = 331 -0.6*15 = 322 m/s$ relative index of refraction $n_r = n_1/n_2 = v_1/v_2 = 343/322 =1.0652$

law of refraction $sin(i)/sin(r) = n_r$

$sin(r) = sin(i)/n_r = sin(20)/1.0652 =0.321$

$r =18.728 degree$

8) A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 6.8 apart. How far away did the impact occur? 


First approximation

The speed of sound in concrete is 10 times bigger than the speed of sound in air. Therefore we can consider the propagation through the concrete near instantaneous

$d = time*v(air) = 343*6.8 = 2332.4 m$

Second approximation:

let $d$ be the distance

time for the sound in concrete to reach us is $t_1 = d/v_1$

time for sound in air to reach us is $t_2 = d/v_2$ (since $v_2$ is 10 times smaller than $v_1$ the distance should be about the same)

$t = t_1-t_2$

$d/v_1 -d/v_2 = t$

$d(v_2-v_1)/(v_1*v_2) =t$

$d = t*v_1*v_2/(v_2-v_1) = 6.8*343*3000/(3000-343) =2633.49 m$

The second approximation is better than the first approximation

9) A 4.3 dB sound wave strikes an eardrum whose area is 6.0×10−5.  How much energy is absorbed by the eardrum per second?  At this rate, how long would it take your eardrum to receive a total energy of 7.3 ? a)


$dB = 10*log(I/I_0)$ where $I_0 = 10^{-12} W/m^2$

$4.3=10*log(I/I0) $

$I/I_0 = 10^4.3 =19952.6$

$I = 19952.6*10^{-12} =1.995*10^{-8} W/m^2$

For a total area of $S =6*10^{-5} m^2$ we have a power of

$P = S*I = 1.995*10^{-8}*6*10^{-5} = 11.97*10^-{13} W=1.197*10^{-12} W$


For a total energy of 7.3 Joules it takes

$P = E/time$

$time = Energy/power =7.3/1.197*10^{-12} =6.1*10^{11} seconds = 19342.97 years$

10) In one of the original Doppler experiments, a tuba was played on a moving flat train car at a frequency of 52 , and a second identical tuba played the same tone while at rest in the railway station. What beat frequency was heard if the train car approached the station at a speed of 14.3 ?


the relation for frequencies in Doppler effect is

$F = (c+v_r)/(c+v_s) * F_0$

where c is the speed of sound in air $c = 343 m/s$

$v_r$ is the speed of the receiver positive if the receiver moves toward the source $v_r = 0$

$v_s$ is the speed of the source positive if the source moves away from the receiver $v_s =-14.3 m/s$

$F = 52*(343/(343-14.3) )=54.26 Hz$

the frequency of the beats is $F_b = 54.26 -52 =2.26 Hz$

the frequency of the beats is about the same for different air temperatures (when the speed of sound varies) . For $c = 333 m/s$ we have $F_b =2.33 Hz$

11) A uniform narrow tube 1.69 long is open at both ends. It resonates at two successive harmonics of frequencies 320 and 384.  What is the fundamental frequency?  What is the speed of sound in the gas in the tube? a)

open at both ends means maxim of standing wave at both ends. This means $length = \lambda/2$ or $length = \lambda$ and in general $length = (n+1)*\lambda/2$

the fundamental is when $length = \lambda/2$

$\lambda = 2*L/(n+1)$

but $\lambda = c/F$

$c/F = 2*L/(n+1)$

$c/F_0 =2*L/n$

$F_0/F = n/(n+1)$

$ 320/384 =n/(n+1)$

$5/6 =n/(n+1)$ implies $n=5$

the fundamental frequency is $F = 320/n = 320/5 = 64 Hz$

at the fundamental we have

$\lambda/2 = 1.69 \lambda =3.38 m$

but $\lambda = C/F$ where $C$ is the speed of sound

$C = F*\lambda = 3.38*64 =216.32 m/s$

12) The frequency of a steam train whistle as it approaches you is 527 . After it passes you, its frequency is measured as 457. How fast was the train moving (assume constant velocity)?a)

see problem 15.10)

$F = F_0* (c+v_r)/(c+v_s)$

$v_r = 0$,

$F= F0 *C/(C-V_s)$

$F= F0*C/(C+V_s)$

$527/457 = (C+V_s)/(C-V_s)$

$C+V_s =1.153*(C-V_s)$

$0.153*C = 2.153*V_s$

$V_s =0.0711*C = 24.4 m/s$ where $c = 343 m/s$ is the speed of sound in the air

13) Some rear view mirrors produce images of cars behind you that are smaller than they would be if the mirror were flat. What is a mirror’s radius of curvature if cars 18.0 away appear their normal size?

In the case of a concave mirror one has

$1/p_1 +1/p_2 =1/f$

where $p_1$ is the position of the object, positive to the left of the mirror

$p_2$ is the position of the image, positive to the right of the mirror

$1/p_2 =1/f-1/p_1$

$p_2 = f*p1/(P_1-f) > 0$

magnification is $Y_2/Y_1 = -P_2/P_1 = f/(P_1-f)

$0.41 = -f/(18-f)$

$0.41*(18-f) =-f$

$0.38 -0.41*f =-f$

$-0.59*f =7.38$

$f = – 12.50 m$

focal distance $f = R_/2 R$ is the curvature radius

$R = 2*f =2*12.50= -25 m$

14) A fisherman’s scale stretches 3.5 when a 2.1 fish hangs from it.

 What will be the amplitude and frequency of vibration if the fish is pulled down 2.3 more and released so that it vibrates up and down?  Find the Frequency

The amplitude is 2.3 cm exactly the difference in the stretches.

The elastic constant is $k = F/d_1 = m*g/d_1 = 2.1*9.81/0.035 =588.6 N/m$

The period of oscillation for an elastic medium is $T = 2*\pi*\sqrt{m/k} = 2*\pi*\sqrt{2.1/588.6} = 0.375 sec

Frequency is $F = 1/T = 2.66 Hz$

15) An organ pipe is 119 long.

What are the fundamental and first three audible overtones if the pipe is closed at one end?

What are the fundamental and first three audible overtones if the pipe is open at both ends?


closed at one end and open at the other end it means $L = \lambda/4$ ($\lambda$ is the wavelength)

$\lambda = 4*L =4*1.19 =4.76 m$

$\lambda = C/F_0$

$F0 = C/\lambda = 343/4.76 = 72.06 Hz$

first overtone is when $L = \lambda/4 +\lambda/2 = 3*\lambda/4$

$\lambda = (4/3) * L =1.5867 m$

$\lambda = C/F_1$

$F_1 = C/\lambda = 343/1.5867 =216.18 Hz$

second overtone is when $L = \lambda/4 +\lambda = 5*\lambda/4$

$\lambda = (4/5 )*L =0.952 m$

$F_2 =C/\lambda = 343/0.952 =360.29 Hz$

third overtone is when $L = \lambda/4 +\lambda/2 + \lambda = (7/4) * \lambda$,

$\lambda = (4/7) *L =0.68 m$

$F3 =C/\lambda = 343/0.68 =504.41 Hz$


open at both ends it means maxim of vibration at both ends

$length = (n+1)*\lambda/2$

$n =0$

$L = \lambda/2$

$\lambda = 2*L = 2*1.19 = 2.38 m$

$\lambda = C/F_0$

$F_0 = C/\lambda = 343/2.38 =144.11 Hz$

first over tome is for $n=1$

$L = \lambda = 1.19 m$

$F_1 = C/\lambda = 343/1.19 =288.23 Hz$

second overtone is for $n=2$

$L = (3/2)*\lambda$,

$\lambda =0.793 m$

$F_2 = C/\lambda =343/0.793 =432.35 Hz$

third overtone is for $n =3$

$L = 2*\lambda$,

$\lambda = L/2 =1.19/2 = 0.595 m$

$F_3 = c/\lambda = 343/0.595 =576.47 Hz$

16) A bungee jumper with mass 63.5 jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting a low point eight more times in 38.6 . He finally comes to rest 29.5 below the level of the bridge. Calculate the spring stiffness constant and the unstretched length of the bungee cord.

Period of oscillation is $T = 38.6/8 =4.825 sec$

$T = 2*\pi*\sqrt{m/K}$ is the mass $k$ is the elastic constant

$\sqrt{m/k} = T/(2\pi) =4.825 / (2\pi) =0.768$

$m/k = 0.590$

$k = 0.590*63.5 =37.45 N/m$


$x = F/k = m*g/k = 63.5*9.81/37.45 =16.63 m$

the unstreched length is $L =29.5 -16.63 =12.86 m$

17) The springs of a 1400 car compress 4.0when its 73 driver gets into the driver’s seat. If the car goes over a bump, what will be the frequency of vibrations?

$k = F/x = m*g/x =73*9.81/0.004 = 179032.5 N/m$

total mass $m = 1400+73 =1473 kg$

$T = 2*\pi*\sqrt{m/K} = 2*\pi*\sqrt{1473/179032.5} =0.570 seconds$

$F = 1/T = 1.754 sec$

18) It takes a force of 73.0 to compress the spring of a toy popgun 0.225 to “load” a 0.190 ball. With what speed will the ball leave the gun?

Potential energy stored is $E = k*x^2 /2$

$k* =F/x =73/0.225 =324.4 N/m$

$E = 324.4*0.225^2 /2=8.2125 J$

Potential energy is equal to the kinetic energy of the ball

$m*v^2/2 = 8.2125$

$v^2 =2*8.2125/0.190 =86.44 m^2/s^2$ thus $v =9.3 m/s$

19) A pendulum makes 29 vibrations in exactly 60. What is its period? What is its frequency?

$T = time/ N = 60/29 =2.07 seconds$

$F = 1/T = 0.483 Hz$

20) Speakers are at opposite ends of a railroad car as it moves past a stationary observer at 10.0 , as shown in Figure(Figure 1) . If the speakers have identical sound frequencies of 219 Hz:

What is the beat frequency heard by the observer when he listens from the position A, in front of the car?

What is the beat frequency heard by the observer when he is between the speakers, at B?

What is the beat frequency heard by the observer when he hears the speakers after they have passed him, at C? A)

when he is in front of the card the frequency of the beats is Fb =0 since both speakers are moving toward the observer and have the same shift in frequency.


when the observer is between the speakers the shift in frequencies of the speakers are

$F_1 =F_0 (c/(c+v_s)) = 219 *(343/(343+10) =212.79 Hz$

$F_2 = F_0 (C/C-V_s)) =219 *343/(343-10) =225.57 Hz$

Frequency of beats is

$F_b = F_2-F_1 =225.57 -212.79 =12.78 Hz$


when he is in position C also the frequency of the beats is $F_b =0$ since both speakers are moving away from the observer and have the same shift in frequency.

21) An aquarium filled with water has flat glass sides whose index of refraction is 1.56. A beam of light from outside the aquarium strikes the glass at a 43.5.  What is the angle of this light ray when it enters the glass?

$sin(i)/sin(r) = n(r)$

$sin(r) = sin(i)/n(r) = sin(43.5)/1.56 =0.441$

$r =26.18 degree$

22) When you slosh the water back and forth in a tub at just the right frequency, the water alternately rises and falls at each end, remaining relatively calm at the center. Suppose the frequency to produce such a standing wave in a 60 -wide tub is 0.75 . What is the speed of the water wave?

you have at both ends maxim of standing waves. $L = \lambda/2$

$\lambda = 2*L = 2*0.6 =1.2 m$

$\lambda = C/F$

$C = F*\lambda = 0.75*1.2 =0.9 m/s$

23) A dentist wants a small mirror that, when 1.90from a tooth, will produce a (upright) image.  What must its radius of curvature be?

In the case of the convex mirror one has the equation

$1/p1-1/p2 =-1/f$

where $p_1$ is the position of the object, positive if the object is to the left of the mirror

$p_2$ is the position of the image, positive if the image is to the right of the mirror,

and $f$ is the focal distance $F= R/2$ ($R$ is the curvature)

The magnification is $Y_2/Y_1 =p_2/p_1 =5$

$p_1 = 1.9 cm$

$p_2 = 5*p_1 =5*1.9 = 9.5 cm$

$1/1.9 – 1/9.5 =-1/f$

$1/f = 0.42$

$f = 2.375 cm$

$f = R/2$

$R = 2*f =-2*2.375 = 4.75 cm$

24) The magnification of a convex mirror is times for objects 6.3 from the mirror. What is the focal length of this mirror?

see problem above.

$1/p1-1/p2 =-1/f$

where $p_1$ is the position of the object, positive if the object is to the left of the mirror

$p_2$ is the position of the image, positive if the image is to the right of the mirror,

and f is the focal distance

The magnification is $Y_2/Y_1 =p_2/p_1 = -0.55$

$p_1 =6.3 m$

$p_2 =-6.3*0.55 = -3.465 m$

$1/6.3 +1/3.465 =-1/f$

$1/f =0.4473$

$f = 2.235 m$