Ph12a_s8_2009

# Ph12a_s8_2009 - Ph12a Solution Set 8 6.14 The dispersion...

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: Ph12a Solution Set 8 November 24, 2009 6.14 The dispersion relation for a system of coupled pendula is given by eq. 90 of section 2.4: ω 2 ( k ) = g l + 4 K M sin 2 ka 2 The lower cutoff frequency is p g l corresponding to k = 0 while the upper cutoff frequency is given by q g l + 4 K M corresponding to k = π a (see figure p.86 of Crawford). We can now evaluate the group velocity: v g = dω dk = 1 2 ω dω 2 dk v g = 1 2 ω 8 K M sin ka 2 cos ka 2 plugging in k = 0 for the lower cutoff frequency we find the that sin ka 2 factor is zero, and thus v g = 0. Taking k = π a the cos ka 2 term becomes zero and so the group velosity is zero at the high frequency cutoff as well. The phase velocity at the two frequencies can be found simply from: v φ = ω k at k = 0 this yields an infinite phase velocity. At k = π a we have v φ = a π q g l + 4 K M A nice graph of the dispersion relation in question is on page 86 of the text. One can read off the group velocity at a particular frequency ω by looking the slope of the tangent to the graph at the point ( k,ω ). The phase velocity can be read off of the diagram by calculating the slope of the line connecting the point representing the frequency ( k,ω ) with the origin. 1 Ph12a Solutions Problem Set 8 November 24, 2009 6.18 (a) The index of refraction is given by: n = c/v φ = ck/ω . So: n 2 = c 2 k 2 ω 2 = 1 + ω 2 p ω 2 o- ω 2 . n 2 is an increasing function of ω on 0 ≤ ω < ω o and ω o < ω < ∞ ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 6

Ph12a_s8_2009 - Ph12a Solution Set 8 6.14 The dispersion...

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

View Full Document
Ask a homework question - tutors are online