Question

What will be the fundamental frequency and first three overtones for a 26cm long organ at 20°C if it is open ( at 20°C, the speed of sound in air is 343m/s)

Answer 1

The fundamental frequency of the 26cm long organ pipe is 659.62 Hz. The first overtone is 1319.24 Hz and the second overtone is 1978.86 Hz.

Step-by-step explanation:

The fundamental frequency of an open pipe (open at both ends) can be calculated using the formula:

f = (v / λ)

where f is the fundamental frequency, v is the speed of sound, and λ is the wavelength.

In this case, the length of the organ pipe is given as 26 cm, which is equal to 0.26 m. We need to determine the wavelength of the fundamental frequency in order to calculate the frequency.

Since the pipe is open at both ends, the fundamental frequency corresponds to the first harmonic. The wavelength of the first harmonic in an open pipe is equal to twice the length of the pipe:

λ = 2 * L

Substituting the values into the equation, we have:

λ = 2 * 0.26 m = 0.52 m

Now we can calculate the fundamental frequency:

f = (343 m/s / 0.52 m) = 659.62 Hz

The first overtone is the second harmonic, which is twice the frequency of the fundamental frequency:

First overtone = 2 * 659.62 Hz = 1319.24 Hz

The second overtone is the third harmonic, which is three times the frequency of the fundamental frequency:

Second overtone = 3 * 659.62 Hz = 1978.86 Hz

Answer 2

Step-by-step explanation:

The frequency of an organ pipe if it is open is given by :

`f=(nv)/(2l)`

v is speed of sound in air is 343 m/s at 20°C

For fundamental frequency, n = 1

`f=(1* 343)/(2* 0.26)\\\\f=659.61\ Hz`

First overtone frequency,

`f_1=2f\\\\f_1=2* 659.61\\\\f_1=1319.22\ Hz`

Second overtone frequency,

`f_2=3f\\\\f_2=3* 659.61\\\\f_2=1978.83\ Hz`

Third overtone frequency

`f_3=4f\\\\f_3=4* 659.61\\\\f_3=2638.44\ Hz`

Hence, this is the required solution.