OLDHW35 - oldhomewk 35 – PAPAGEORGE MATT – Due 4:00 am...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: oldhomewk 35 – PAPAGEORGE, MATT – Due: Apr 28 2008, 4:00 am 1 Question 1, chap 16, sect 4. part 1 of 1 10 points Two guitar strings, of equal length and lin- ear density, are tuned such that the second harmonic of the first string has the same fre- quency as the third harmonic of the second string. The tension of the first string is 210 N. Calculate the tension of the second string. Correct answer: 93 . 3333 N (tolerance ± 1 %). Explanation: The frequency of the n th harmonic of a string is f = n 2 l radicalBigg F μ . (This arises from the requirement that an integer number of half-wavelengths must fill the string in order to keep the end-points of the string fixed.) For two strings of equal lengths and linear densities, we get f 2 f 1 = n 2 n 1 radicalbigg F 2 F 1 . In this case f 2 = f 1 , so F 2 = F 1 parenleftbigg n 1 n 2 parenrightbigg 2 = 93 . 3333 N . Question 2, chap 16, sect 4. part 1 of 1 10 points Two sound sources radiating in phase at a frequency of 540 Hz interfere such that max- ima are heard at angles of 0 ◦ and 26 ◦ from a line perpendicular to that joining the two sources. The volicity of sound in air is 340 m/s. y L d S 1 S 2 θ listening direction θ δ Find the separation between the two sources if the velocity of sound is 340 m / s. Correct answer: 1 . 43629 m (tolerance ± 1 %). Explanation: Let : f = 540 Hz , v = 340 m / s , θ = 0 ◦ , and θ = 26 ◦ . r 2 r 1 y L d S 1 S 2 θ = ta n − 1 parenleftBig y L parenrightBig listening direction δ ≈ d sin θ ≈ r 2- r 1 P O negationslash S 2 QS 1 ≈ 90 ◦ Q 2 π or 360 ◦ φ δ λ = 500 1- 1 ◦ 90 ◦ 180 ◦ 270 ◦ 360 ◦ π/ 2 π 3 π/ 2 2 π φ 2 π = δ λ = d sin θ λ = d λ y radicalbig L 2 + y 2 ≈ d λ y L Because a maximum is heard at 0 ◦ and the sources are in phase, we can conclude that the path difference is 0. Because the next maximum is heard at 26 ◦ , the path difference to that position must be one wavelength. oldhomewk 35 – PAPAGEORGE, MATT – Due: Apr 28 2008, 4:00 am 2 sin θ = Δ s d d = λ sin θ = v f sin θ = 340 m / s (540 Hz) sin26 ◦ = 1 . 43629 m . For additional intensity maxima, d sin θ m = mλ, m = 2 , 3 , 4 , ... θ m = sin − 1 parenleftbigg mλ d parenrightbigg , m = 2 , 3 , 4 , ... The next angle is for m = 2: θ m =2 = sin − 1 parenleftbigg mλ d parenrightbigg = sin − 1 parenleftbigg mv f d parenrightbigg = sin − 1 bracketleftbigg 2 (340 m / s) (540 Hz) (1 . 43629 m) bracketrightbigg . Question 3, chap 16, sect 4. part 1 of 1 10 points A sinusoidal wave in a rope is described by the wave function y = A sin( k x + ω t ) , where x and y are in meters, t is in seconds, A = 0 . 12 m, k = 0 . 68 m − 1 , and ω = 24 Hz....
View Full Document

{[ snackBarMessage ]}

Page1 / 9

OLDHW35 - oldhomewk 35 – PAPAGEORGE MATT – Due 4:00 am...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online