Question

The frequency driving an open-closed tube is steadily raised. Resonances are heard at 500 hz and 700 hz with none in between. What is the fundamental and next highest harmonic frequency?

Answer 1

Answer:

Explanation:

Frequency for a Open-closed tube is given by

$f=(2n+1)\frac{V}{4L}$

n=0 for Fundamental frequency

$f_0=\frac{V}{4L}$

the Next higher frequency

$f_1=(2\times 1+1)\frac{V}{4L}=\frac{3V}{4L}$

$f_1=3f_0$

also

$\frac{(2n+1)\frac{V}{4L}}{(2n+3)\frac{V}{4L}}=\frac{500}{700}$

$\frac{2n+1}{2n+3}=\frac{5}{7}$

$14n+7=10n+15$

$n=2$

thus $\frac{V}{4L}=100 Hz$

Therefore basic fundamental frequency is 100 Hz

and next highest harmonic frequency is

$f_1=3f_0=3\times 100=300 Hz$