见 L1=(2n+1)=0.34m L2-Z1==0.69m L2=[2(n+1)+1]=1.3m 故x=2(L2-L1)=138m 声速=v=138×2485=343ms 1.38m 因L1=(2n+1),=(2n+1) =0.34m LI 得n=0→ 36 0.34 m 4 ( 2 1 ) 1 = + =  L n   1.0 3 m 4 2 ( 1 ) 1 2 = + + =  L n 故 0.69m 2 2 − 1 = =  L L  = 2 ( L2 − L1 ) = 1.38 m 声速 u =   = 1.38 248.5 = 343m/s 因得 n = 0 4 1  L = 43 2  L = L 1 L 2 0.34 m 4 1.38 m (2 1) 4 (2 1) L1 = n + = n + = 