hw12_solutions

# hw12_solutions - Physics 221 Homework 12 Spring 2016...

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

Physics 221 Name/Section: Homework 12 Spring 2016 12.1 A wire is made out of a special steel alloy with 11 2.00 10 Pa Y = × , density ρ = 7800 kg/m 3 , radius 1.00 mm, and coefficient of linear thermal expansion α = 2.50 × 10 5 K 1 . The initial temperature of the wire is 20°C. As a constant tension of 3.53 x 10 4 N is applied to the wire, the wire is measured to be exactly 60.0 cm long. (a) What is the fundamental frequency of the wire? Assume the speed of sound is 344 m/s. (b) Where should you pluck the wire to achieve the third harmonic (second overtone) without also exciting the fundamental mode, and what is its frequency? (c) If the wire is heated to the melting point of steel (1363 o C), by how much does the fundamental frequency change? Solution: (a) The fundamental frequency for a wire fixed at both ends is given by: f 1 = v 2 L = 1 2 L T μ The mass per unit length is μ = m L = ρ V L = ρ AL L = ρπ r 2 = (7800)(3.14)(.001) 2 = 2.45 × 10 2 kg/m so the fundamental frequency is f 1 = v 2 L = 1 2 L F μ = 1 2(0.600) 3.53 × 10 4 2.45 × 10 2 = 1000 Hz. The nth harmonic is f n = n f 1 , so f 3 = (3) f 1 = 3000 Hz. The nodes for the third harmonic are 1/3 and 2/3 of the way from one end to the other, and the antinodes are at 1/6, 3/6, and 5/6 of the way from one end to the other. We must pluck an antinode to excite a given frequency. Plucking at the middle also excites the fundamental frequency, so we need to pluck at L/6 or 5L/6 (10 cm from either end) to preferentially excite just the third harmonic. (c) The initial temperature is 20°C and the final temperature is 1363 o C, so Δ L = L α Δ T = (0.600)(2.50 × 10 5 )(1343) = 0.0201 m Note that the density changes as the length of the wire changes, but the mass remains the same. The mass before heating is m = μ L = (2.45 × 10 2 )(0.600) = 1.47 × 10 2 kg

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

View Full Document
Physics 221 Name/Section: Homework 12 Spring 2016 Then the new fundamental frequency is f = v 2 L = 1 2 L T μ = 1 2 L TL m = 1 2 T mL = 1 2 3.53 × 10 4 (.0147)(0.6201) = 984 Hz 12.2 (a) In which of these scenarios do you hear a higher pitched sound? Explain. (i) A train with its horn sounding is stationary and you move directly towards it with a speed v 0 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern