March 25, 2014

Musical Instrumentsand

Geometric Optics

This is how the picture wastaken. Two plane mirrorsWere facing each other withThe candle in between.

Announcements & Reminders• To the teacher: Turn on the recording!•  

• To students: • About equipment needed for L17

B = 2683 Hzf = B/2 = 427.0 HzFrequency printed on tuning fork = 426. 6 Hz

L149, Problem 1

B = 1608 Hzf = B/2 = 255.9 HzFrequency printed on tuning fork = 256 Hz

n1 = 256 Hz

n2 = 2x256 Hz n6 = 6x256 Hz

L149, Problem 2

B = 9498 Hzf = B/2 = 1512 Hz

vsound = 344 m/s at 21 oC= v/f = 0.23 m

The wavelength is the length of the pipe. This requires the waveform shown to the right. n = 2

A N A N A

L149, Problem 3

B = 1.502E4 Hzfn = B/2 = 2391 Hz

From the last problem,f2 = 1512 Hz.

Thus, f1 = f2/2 = 756 Hz.

n = 6

n = 2391 / 756 Hz = 3.16

This must be the3rd harmonic.

L149, Problem 4

B = 2682 Hzfn = B/2 = 427 Hz

From the last problem,f1 = 756 Hz.

L149, Problem 5

The frequency is 0.56f1. Thisis approximately an octavelower and may be produced byclosing the end of the pipe asseen in P211.

Why isn’t the frequency exactly an octave lower than the fundamental of the open pipe? The effective lengths of the open and closed pipes may be different. For a closed pipe with a fundamental of 427 Hz, the wavelength is= (344 m/s)/(427 Hz) = 0.806 m. For the fundamental of a closed pipe Lc = /4. In this case, that gives Lc = 0.203 m. A smaller length would make sense, since closing the end prevents the waveform from extending beyond the end of the pipe.

Harmonics of a Longer Toy Flute

f1 = 546 Hz

f2 = 1087 Hz (2f1 = 1092 Hz)

n = 2 n = 4 n = 6 n = 8

Waveforms and Frequency Spectra of Various Instruments

waveform by Justin Luo

waveform by Ashwin Monian

waveform by Hunter Denham

waveform by Jennie Cunningham

waveform by Hayden Rudd

waveform by Jesse Sykes

waveform by Will King

waveform by Will King

waveform by Charles Zhao

waveform by Arjun Adhia

waveform by Drew Smith

waveform by Trey Faddis

waveform by Param Sidhu

waveform by Hunter Denham

waveform by Kayla Howes

waveform by Michael Schroeder

waveform by Jesse Sykes

Mirror Diagnostic Questions

The mirror is grayed out sothat you can’t see whetherit’s plane, convex, or concave.The object is the black, uprightarrow, and the image is thegray, inverted arrow.

1.Is the mirror plane or curved?

The mirror is grayed out sothat you can’t see whetherit’s plane, convex, or concave.The object is the black, uprightarrow, and the image is thegray, inverted arrow.

1.The mirror is curved. (The object and image are different sizes.)

2.Is the image real or virtual?

The mirror is grayed out sothat you can’t see whetherit’s plane, convex, or concave.The object is the black, uprightarrow, and the image is thegray, inverted arrow.

1.The mirror is curved. (The object and image are different sizes.)

2.The image is real. (Light rays actually converge at the position of the image.)

3.Is the mirror convex or concave?

The mirror is grayed out sothat you can’t see whetherit’s plane, convex, or concave.The object is the black, uprightarrow, and the image is thegray, inverted arrow.

1.The mirror is curved.

2.The image is real.

3.The mirror is concave. (A convex mirror won’t produce a real image of a real object.)

Everything is revealed here.

o = object distance i = image distance f = focal lengthM = magnification

Where should the object beplaced so that the image isinverted and the same sizeas the object?

Where should the object beplaced so that the image isinverted and the same sizeas the object?

The object must be placedat the center of curvatureof the mirror. This is twicethe focal length.

1. Is the image real or virtual?2. What kind of mirror is it?

1. The image is virtual. (It’s behind the mirror.)

2. The mirror is concave. (The image is larger than the object.)

A convex mirror producesvirtual images smallerthan the object.

Hence the phrase, “Objects in the mirror are closer thanthey appear.”

1. Is the image real or virtual?2. What kind of mirror is it?

1. The image is virtual.2. The mirror is plane.

1. Is the image real or virtual?2. What is the image distance?3. What is the magnification?

1. The image is real. 2. The image distance is found using 1/i = 1/f – 1/o. The object distance and focal length are read from the scale. o = 12.0 cm and f = 10.0 cm. This gives i = 60.0 cm.

3. The magnification is –i/o = -5.00. This also equals the ratio of image height to object height. The negative sign indicates the inversion of the image.

1. The image is real. 2. Compare I = 60.0 cm found in the previous slide to 59.9 cm above.

3. Compare M = -5.00 found in the previous slide to -4.99 above.