Homework 5 solutions

# Homework 5 solutions - WebAssign Problems 1 Two pulses...

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

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

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

Unformatted text preview: WebAssign Problems 1. Two pulses traveling on the same string are described by y 1 = 5 3 x- 4 t ( ) 2 + 2 and y 2 =- 5 3 x + 4 t- 6 ( ) 2 + 2 (a) In which direction does each pulse travel? (b) At what time do the two cancel everywhere? (c) At what point do the two pulses always cancel? Solution: (a) ( ) 1 y f x vt =- , so wave 1 travels in the direction x + ( ) 2 y f x vt = + , so wave 2 travels in the direction x- (b) To cancel, 1 2 y y + = : ( ) ( ) 2 2 5 5 3 4 2 3 4 6 2 x t x t + =- + +- + ( ) ( ) ( ) 2 2 3 4 3 4 6 3 4 3 4 6 x t x t x t x t- = +-- = ± +- for the positive root, 8 6 t = 0.750 s t = (at 0.750 s t = , the waves cancel everywhere) (c) for the negative root, 6 6 x = 1.00 m x = (at 1.00 m x = , the waves cancel always) 2. Two identical sinusoidal waves with wavelengths of 3.00 m travel in the same direction at a speed of 2.00 m/s. The second wave originates from the same point as the first, but at a later time. Determine the minimum possible time interval between the starting moments of the two waves if the amplitude of the resultant wave is the same as that of each of the two initial waves. Solution: 2 cos 2 A A φ = so 1 1 cos 60.0 2 2 3 φ π- = = ° = Thus, the phase difference is 2 120 3 π φ = ° = This phase difference results if the time delay is 1 3 3 3 T f v λ = = Time delay ( ) 3.00 m 0.500 s 3 2.00 m s = = 3. Two loudspeakers are placed on a wall 2.00 m apart. A listener stands 3.00 m from the wall directly in front of one of the speakers. A single oscillator is driving the speakers at a frequency of 300 Hz. (a) What is the phase difference between the two waves when they reach the observer? (b) What If? What is the frequency closest to 300 Hz to which the oscillator may be adjusted such that the observer hears minimal sound? Solution: (a) 9.00 4.00 3.00 13 3.00 0.606 m x Δ = +- =- = The wavelength is 343 m s 1.14 m 300 H z v f λ = = = Thus, 0.606 0.530 of a w ave 1.14 x λ Δ = = , or ( ) 2 0.530 3.33 rad φ π Δ = = (b) For destructive interference, we want...
View Full Document

## This note was uploaded on 11/22/2010 for the course PHYSICS Physics 7C taught by Professor Ilyakrivorotov during the Winter '10 term at UC Irvine.

### Page1 / 7

Homework 5 solutions - WebAssign Problems 1 Two pulses...

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

View Full Document
Ask a homework question - tutors are online