彈性索的一端固定在振動器上，另一端固定在支架上

假設 12 Hz 時, 駐波圖形有 N 個圈 (loop), 則在 16 Hz 時, 駐波圖形有 N + 1 個圈,

所以:

12 Hz 時:

Nλ/2 = 1 (每個圈表示 λ/2 的距離)

Nv/(12 x 2) = 1 (以 v = fλ 計)

N = 24/v ... (1)

16 Hz 時:

(N + 1)λ'/2 = 1 (每個圈表示 λ'/2 的距離)

(N + 1)v/(16 x 2) = 1 (以 v = fλ' 計)

N + 1 = 32/v ... (2)

(2) - (1):

1 = 8/v

v = 8 m/s

Assume there are n "loops" when the vibrator is operated at 12 Hz, hence length of one loop = 1/n
Since one wavelength equals to the length of two loops, wavelength 入 = 2/n
But 入 = v/12, where v is the speed of wave on the string
Therefore, 入 = v/12 = 2/n
i.e. n = 24/v

Similarly, when the string vibrates at 16 Hz, there are (n+1) loops. Hence,
new wavelrngth 入' = v/16 = 2/(n+1)
Thus, v/16 = 2/[(24/v) + 1]
solve for v gives v = 8 m/s