Question

What is the fundamental frequency of a 0.003 kg steel piano wire of length 1.3 m and under a tension of 2030 N? Answer in units of Hz. 005 (part 2 of 2) 10.0 points What is the fundamental frequency of an organ pipe 1.38 m in length, closed at the bottom and open at the top? The speed of sound is 340 m/s. Answer in units of Hz.

Answer 1

a) 277.5 Hz

b) 61.6 Hz

Explanation:

a)

Stationary waves are the waves produced on a string: these are waves that do not propagate through space, since they travel only back and forth along the string.

These waves can have different frequencies, depending in how many segments of the string they vibrate.

The frequency of the fundamental mode of vibration (the one having only two nodes at the ends of the string) is called fundamental frequency, and it is given by:

$f_1=\frac{1}{2L}\sqrt{\frac{T}{m/L}}$

where

L is the length of the spring

T is the tension

m is the mass of the string

For the steel piano wire in this problem:

T = 2030 N

L = 1.3 m

m = 0.003 kg

Therefore, the fundamental frequency is

$f_1=\frac{1}{2(1.3)}\sqrt{\frac{2030}{(0.003)/(1.3)}}=277.5 Hz$

b)

An organ pipe is a closed-air tube, which is open at one end and close at the other end.

For a closed-open air tube, the wavelength of the fundamental mode of vibration is equal to 4 times the length of the tube:

$\lambda_1 = 4 L$

In this case, the length of the tube is

L = 1.38 m

So the fundamental wavelength is

$\lambda_1 = 4(1.38)=5.52 m$

The relationship between frequency and wavelength for a sound wave is

$f_1=\frac{v}{\lambda_1}$

where in this case:

v = 340 m/s is the speed of sound

$\lambda_1 = 5.52 m$ is the fundamental wavelength

Solving for f1, we find the fundamental frequency:

$f_1=\frac{340}{5.52}=61.6 Hz$