 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2022/paper-2/section-b/b6/b/ii/solution

For a fixed transverse normal mode, define its cutoff angular frequency

ω_{c, n} = \frac{nπ v}{b}, ω^{2} = ω_{c, n}^{2} + v^{2} k_{x}^{2} .

The longitudinal phase velocity and group velocity, for propagation in

+ x

, are

v_{p} = \frac{ω}{k _{x}} = \frac{v}{1 - ( ω _{c, n} / ω ) ^{2}}, v_{g} = \frac{d ω}{d k _{x}}_{n} = \frac{v ^{2} k _{x}}{ω} = v 1 - (ω_{c, n} / ω)^{2} .

Thus

v_{p} v_{g} = v^{2}

v_{p} > v

and

v_{g} < v

. The large longitudinal phase velocity reflects the geometry of the transverse standing wave and does not imply faster transport of energy than the membrane wave speed.

n

f_{c, n} = \frac{ω _{c, n}}{2 π} = \frac{n v}{2 b} .

At cutoff

k_{x} = 0

and

v_{g} = 0

; the longitudinal phase velocity has no finite value. Below cutoff,

k_{x} = i γ_{n}, γ_{n} = (nπ / b)^{2} - ω^{2} / v^{2},

and a bounded field excited at

x = 0

decays as

e^{- γ_{n} x}

. It is an evanescent wave, rather than a freely propagating mode carrying net time-averaged energy flux down an indefinitely long uniform strip. A finite section can still transmit through coupling of evanescent fields.

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