HW 33 - Morby Grant Homework 33 Due noon Inst Drummond This...

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

Morby, Grant – Homework 33 – Due: Apr 26 2006, noon – Inst: Drummond 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A transverse wave in a wire has form y ( x, t ) = (1 . 6 cm) × sin h (5 . 08 m - 1 ) × x - (7110 s - 1 ) × t i The wire has linear density μ = 7 . 59 g / m. What is its tension? Correct answer: 14868 N. Explanation: A sinusoidal transverse wave of the general form y ( x, t ) = A × sin( kx - ωt ) travels in the positive x direction at speed v = ω k . For the wave in question, this speed is v = 7110 s - 1 5 . 08 m - 1 = 1399 . 61 m / s . For a transverse wave in a wire, this speed fol- lows from the wire’s tension and linear density according to v = s F μ , hence F = μ × v 2 = 14868 N . 002 (part 1 of 1) 10 points A harmonic wave y = A sin[ k x - ω t - φ ] , where A = 1 meter, k has units of m - 1 , ω has units of s - 1 , and φ has units of radians, is plotted in the diagram below. +1 - 1 A (meters) x (meters) 10 m 20 m 30 m At the time t = 0 Which wave function corresponds best to the diagram? 1. y = A sin 2 π 15 m x - ω t - 4 π 3 ¶‚ 2. y = A sin 2 π 3 m x - ω t - 2 π 3 ¶‚ 3. y = A sin 2 π 9 m x - ω t - 5 π 3 ¶‚ 4. y = A sin 2 π 9 m x - ω t - 2 π 3 ¶‚ 5. y = A sin 2 π 3 m x - ω t - 1 π 3 ¶‚ 6. y = A sin 2 π 9 m x - ω t - 4 π 3 ¶‚ 7. y = A sin 2 π 3 m x - ω t - 4 π 3 ¶‚ 8. y = A sin 2 π 15 m x - ω t - 1 π 3 ¶‚ correct 9. y = A sin 2 π 15 m x - ω t - 2 π 3 ¶‚ 10. y = A sin 2 π 15 m x - ω t - 5 π 3 ¶‚ Explanation: From the diagram of the wave function the wave-length λ = 15 m (6 horizontal scale di- visions of 2 . 5 m each, see diagram below). Notice: Since one wave-length is 2 π ra- dians, each horizontal division is 2 π 6 = π 3 radians.
Image of page 1

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

Morby, Grant – Homework 33 – Due: Apr 26 2006, noon – Inst: Drummond 2 The given wave function (sine function with t = 0) y = A sin k x - φ = A sin 2 π λ x - φ = A sin 2 π 15 m x - φ (dark curve in diagram below) is shifted 1 divisions to the right (negative phase shift) of a no-phase-shift sine function y = sin 2 π 15 m x (gray curve in diagram below), therefore φ = 1 π 3 = 1 π 3 radians .
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.
  • Spring '09
  • KLEINMAN
  • Physics, Work, Sin, noon – Inst, superposed waves, Grant – Homework

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern