Question

The lowest-pitch tone to resonate in a pipe of length l that is closed at one end and open at the other end is 200 hz. Which frequencies will not resonate in the same pipe?

Answer 1

The frequency 400 hz is not possible .

Step-by-step explanation:

Given that,

Frequency = 200 hz

Length = l

Suppose, The given frequencies are,

600 Hz, 1000 Hz, 1400 Hz, 1800 Hz and 400 hz.

The possible resonance frequencies are

We need to calculate the fundamental frequency

Using formula of fundamental frequency for pipe

`F=(nv)/(4L)`

Where, n = odd number

Put the value of frequency

`200= (nv)/(4L)`

We need to calculate the first over tone

Using formula of fundamental frequency

n = 3,

`F=(nv)/(4L)`

Put the value into the formula

`F_(2)=3*(v)/(4l)`

`F_(2)=3*200`

`F_(2)=600\ Hz`

We need to calculate the second over tone

Using formula of fundamental frequency

n = 5,

`F=(nv)/(4L)`

Put the value into the formula

`F_(3)=5*200`

`F_(3)=1000\ Hz`

We need to calculate the third over tone

Using formula of fundamental frequency

n = 7,

`F=(nv)/(4L)`

Put the value into the formula

`F_(4)=7*200`

`F_(4)=1400\ Hz`

We need to calculate the fourth over tone

Using formula of fundamental frequency

n = 9,

`F=(nv)/(4L)`

Put the value into the formula

`F_(5)=9*200`

`F_(5)=1800\ Hz`

Hence, The frequency 400 hz is not possible .