(a) The two resonance conditions are
L=(2n1+1)4λ1and L=(2n2+1)4λ2
Since the first frequency is smaller than the second frequency, the first wavelength is larger than
the second wavelength. This means
n1<n2 or n2=n1+1
Together,
(2n1+1)f1v=(2n1+3)f2v
⇒(2n1+1)f2=(2n1+3)f1
⇒(2n1+1)=(2n1+3)f2f1
⇒(2n1+1)=2n1f2f1+3f2f1
⇒2n1(1−f2f1)=3f2f1−1
⇒n1(0.2666)=1.6⇒n1=0.2661.6=6.01
n1=6 and n2=7
The depth of the well is
L=(2n1+1)4λ1
=413×51.87343=21.491m
(b) The number of anti-nodes for n=6 is 7.
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