Test 2 Practice

course Phy 232

Dr. Smith,I am having a very hard time with the practice test 2. I am going out of the country friday and am trying my hardest to get test 1 and 2 done before I go. I was scheduled to take test 1 and 2 tomorrow, however I am having a great deal of trouble with test 2. Can you please help me with test 2 and give me back your feedback? I am going to talk to my proctor tomorrow, and I hope to take test 2 by friday.

Problem Number 1

Show that the function y(x, t) = .43 e^-( 320 t - .75 x)^2 satisfies the wave equation, and give the frequency, wavelength and velocity of the wave. Sketch the wave from x = - 13 to x = 13 at t=0 and at t = .01359.

I do not quite get how this satisfies the wave equation. The wave equation is y = A sin( (2 `pi f) (t - timeLag). So how is the wave equation now have an exp term.

The wave equation relates the second derivative of y with respect to time t and the second derivative of y with respect to position x. The two are proportional, and the proportionality constant is related to the velocity of propagation of the wave. The derivatives are easy. Look up the wave equation in your text, take the appropriate derivatives and get back to me with your results and/or questions.

For example the first derivatives are

dy/dx = 2 pi f A cos( (2 `pi f) (t - timeLag) (treat x as the variable, everything else as constant)

and

dy/dt = -4 pi^2 f^2 A sin( (2 `pi f) (t - timeLag) (treat t as the variable, everything else as constant).

Problem Number 2

A string of length 6 meters is fixed at both ends. It oscillates in its second harmonic with a frequency of 179 Hz and amplitude .29 cm. If its mass is 8 grams, then what tension is it under?

V=sqrt(tension/mass/unitlength)

6 m/179 Hz= sqrt(tension/.08 kg/6m)

Tension=1.49*10^-5 N

Right idea but you don't have the right velocity.

What's the wavelength of the second harmonic?
How do you put wavelength and frequency together to get propagation velocity?

Problem Number 3

If a traveling wave has wavelength 2.7 meters and the period of a cycle of the wave is .071 second, what is the propagation velocity of the wave? What is its frequency?

Velocity=2.7 meters/.071 second=38.03 m/s

Problem Number 4

A sound source with frequency 130 Hz approaches an observer at 95 m/s. If the speed of sound is 340 m/s, then what frequency will be heard by the observer?

fl=((v+vl)/v)*fs=((340+95)/340)*130=166.3 Hz

be sure you know the equations for both observer approaching source and source approaching observer; it is of course best if you know how to derive those equations

Problem Number 5

A candle flame is 1.8 cm high and 12 cm from a circular mirror of radius 3 cm.

• Where will the image of the flame form and how large will it be?

• Will it be real or virtual? Will it be upright or inverted?

• Sketch a ray diagram explaining how the image is formed.

The image will be virtual and upright.

I do not know how to calculate where and how large the candle’s image will be.

Identify object distance and focal distance, then use 1/f = 1/i + 1/o to find image distance.
The focal length of a spherical mirror is half its radius of curvatuve.

The ratio of image size to object size is the same as the ratio of image distance to object distance, as is found from a ray diagram and similar triangles.

Make sure you know how to sketch the ray diagram for a lens or a mirror, showing the principal rays.

Problem Number 6

A string defines the x axis, with the origin at the left end of the string. The string has tension 19 Newtons and mass per unit length is 14 grams / meter. The left-hand end of the string is displaced perpendicular to the string in a cyclical manner in order to create a traveling wave in the string. At clock time t the position of the left-hand end of a long string is y = .54 cm * sin ( ( 9 `pi rad/s) t ).

• What equation describes the displacement a point 16.7 meters down the string as a function of clock time?

• If the position of the left-hand side is x = 0, what is the equation for the displacement as a function of clock time at arbitrary position x?

• University Physics: Show that your equation satisfies the wave equation.

• What is the equation for the shape of the string at clock time t = .042 sec?.

This problem appears in very similar form, but with different numbers, in the Introductory Problem Set. Make sure you know how to solve everything in that problem set. If you have specific questions on any of the solutions, you're welcome to ask.

Problem Number 7

One sound is 3400000 times louder than a sound which measures 40 decibels. What is the decibel level of this sound?

10 log (P2/P1) = 40 dB, P2/P1=10,000

Sound is 3400000 times louder, so 10,000*3400000=3.4*10^10

10 log (P2/P1) = 10 log (3.4*10^10) = 105.3 dB

good

Problem Number 8

A string is under a tension of 18 Newtons and lies along the x axis. ""Beads"" with mass 7.3 grams are located at a spacing of 20 cm along a light but strong string. At a certain instant a certain bead is at y position .0012 meters, while the bead to its right is at y position .001 meters and the bead to its left at y position .0026 meters. At this instant the bead is moving in the y direction at -.1284 m/s. Find the acceleration of the given bead and approximate its velocity .027 seconds later and the distance it will move in this time.

I am having a really hard time with this problem. I do not know where to start.

this problem is not well-covered in the notes, and would be omitted if you didn't get it

Problem Number 9

Find the frequencies of the first four harmonics of a standing wave in a uniform string of length 9 meters which is fixed at both ends under a tension of 35 Newtons, and which has mass 31 grams.

v=sqrt(tension/mass/unitlength)

v=sqrt(35N/.031/9m)=100.8 m/s

freq=velocity/length

harmonic 1, length=.5L=18m, freq=100.8 m/s/18m=5.6 s^-1

harmonic 2, length=2L/2=9m, freq=100.8 m/s/9m=11.2 s^-1

harmonic 3, length=3L/2=6m, freq=100.8 m/s/6m=16.8 s^-1

harmonic 4, length=4L/2=4.5m, freq=100.8 m/s/4.5m=22.4 s^-1