phy standing wave

Since the pipe is closed at one end and open at the other, there is a displacement node at the close-end and a displacement antinode at the open end.

Hence, the lowest frequency occurs when the length of the tube equals to 1/4 of a wavelength.
Thus wavelength of the note = 4 x 0.5 m = 2 m
Frequency = 340/2 Hz = 170 Hz

At the next higher frequency, the length of pipe equals to 3/4 wavelength.
Hence, wavelength = 4 x 0.5/ 3 m = 0.6667 m
Frequency = 340/0.667 Hz = 510 Hz

The answer is option B.

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|<-0.5m->|

Sound wave and the reflected Sound wave superpose to form Standing Wave.

Because frequency f= v/入,

we just have to find the wavelength.

Recall that wavelength of standing wave are quantized,

L = n 入/2

=>入=2L/n

For the lowest freq, that is, for highest lambda,

put n=1 and n=2,

and then you can find lambda.

Then you can find frequency f.

2012-11-30 14:20:51 補充：
At the beginning, the sentence should be:

The sound wave and the reflected sound wave

superpose to form Standing Wave.