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Unformatted text preview: day, October 9, 2013 √ 2 τ µω A 1
2 2 and
since τ2 √
τ k ω = ω = τ µω
v Pave 1√
τ µω A
sin θ = cos θ =
2 2 3d wave, intensity
Intensity = I = energy ﬂux, i.e. energy ﬂow per area,
per time. In 3d, I drops as distance r-squared, since
energy is conserved and area grows as r-squared.
I = P/4π r
I1 /I2 = Wednesday, October 9, 2013 2 22
r2 /r1 E.g. 60W bulb emits P=60W, intensity of
the light drops as 1/r^2, since light spreads out. Play / tune your guitar
Fingers shorten string length,
shorter length = higher frequency. Bass: fatter + longer strings =
Tune: tighten the knobs
(increase tension) to get
Wednesday, October 9, 2013 String waves, modes
( 2 2−
)y (t, x) = 0.
∂x Ends, y (t, 0) = y (t, L) = 0
The wave eqn
+ B.C. solution: n=3 Ends, and nodes, have y(t)=0. n = 1, 2, 3, 4, . . .
mode higher harmonics nπ x
y (t, x) = A sin(
) cos(ωn t)
ωn = vkn = nω1
µL Wednesday, October 9, 2013 =“mode number” 2π
L Tune your guitar! More bass
(lower freq) from fatter or
longer strings. Higher freq.
from more tension. Harmonics
n=2, (1 node)
n=3 (n-1 nodes)
λn = 2L/n ωn = vkn = nω1
Wednesday, October 9, 2013 2π
L ω1 = τπ
ω ≡ 2π f ≡ 2πν Wave B.C.s at the ends:
Two common choices:
Fixed (Dirichlet): y (t, xend ) = 0 Free (Neumann): ∂y
(t, xend ) = 0
∂x Fixed (Dirichlet)
Wednesday, October 9, 2013 Free (Neumann) Free ends case, sol’n:
Suppose free ends at x=0, and x=L
(t, 0) =
(t, L) = 0
y = A cos(
L Fixed ends (few slides ago): this cos was instead sin.
The harmonics are similar in the two cases. Wednesday, October 9, 2013 Free ends harmonics
Slope =0 at ends λ = 2L λ=L λ = 2L/3 ... λn = 2L/n −
fn = Tn 1 = v/λn = nv/2L Wednesday, October 9, 2013 Just for fun: string
theory! Similar wave equation. Different harmonics are
different particles. Known particles are the fundamental harmonic, others would be new particles. Wednesday, October 9, 2013...
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This note was uploaded on 02/11/2014 for the course PHYSICS 2C taught by Professor Hicks during the Fall '09 term at UCSD.
- Fall '09