540_Physics ProblemsTechnical Physics

540_Physics ProblemsTechnical Physics - 542 P18.53...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
542 Superposition and Standing Waves P18.53 For an echo ′= + ff vv s s bg the beat frequency is f b = ′ − . Solving for f b . gives v b s s = 2 when approaching wall. (a) f b = = 256 2133 343 1 33 199 a f af a f . . . Hz beat frequency (b) When he is moving away from the wall, v s changes sign. Solving for v s gives v fv s b b = = = 2 53 4 3 22 5 6 5 338 a f a fa f . m s . *P18.54 Using the 4 and 2 2 3 - foot pipes produces actual frequencies of 131 Hz and 196 Hz and a combination tone at 196 131 65 4 −= Hz Hz . , so this pair supplies the so-called missing fundamental. The 4 and 2-foot pipes produce a combination tone 262 131 131 Hz Hz, so this does not work. The 2 2 3 2 and -foot pipes produce a combination tone at 262 196 65 4 Hz Hz . , so this works. Also, 42 2 3 2 ,, and -foot pipes all playing together produce the 65.4-Hz combination tone. Section 18.8 Non-Sinusoidal Wave Patterns P18.55
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online