Physics DSE Wave MC

Cheuk Ling

2014-03-22 7:18 am

A stretched string is 50cm long and is fixed at both ends as shown below. If the vibrator increases its frequency gradually, two successive stationary waves would be produced without intermediate frequency. Given the speed of the mechanical wave along the string is 50 ms^-1.

Which of the following is the possible frequency combination of the said two successive sound waves?

A. 250 Hz and 350 Hz
B. 450 Hz and 500 Hz
C. 500 Hz and 600 Hz
D. 630 Hz and 680 Hz

Ans: B

唔該有冇人可以解一解比我聽?? ):

回答 (1)

天同

2014-03-22 7:59 am

✔ 最佳答案

A loop produced on a vibrating string = 入/2
where 入 is the wavelength of the wave
hence, for n loops produced on a string of length L, we have
L = n(入/2)
i.e. 入 = 2L/n

Frequency of stationary wave f = v/入 = nv/2L
where v is the speed of wave on the string.

For two successive stationary wave with n and (n+1) loops, we have
f1 = nv/2L
f2 = (n+1)v/2L
where f1 and f2 are the frequencies of the two successive stationary waves.

Thus, f2 - f1 = (n+1)v/2L - nv/2L = v/2L = 50/(2x 0.5) Hz = 50 Hz
This shows that the two successive waves must have frequencies separated by 50 Hz. Thus, we can exlude options A and C.

Since the fundamental frequency (by setting n = 1) = 50/(2x0.5) Hz = 50 Hz
the frequencies increase from 50 Hz in intervals of 50 Hz, i.e. 50 Hz, 100 Hz, 150 Hz, 200 Hz, 250 Hz....and so on.

Therefore, it is obvious that option B is the correct answer, as 630 Hz and 680 Hz are not harmonics.

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