Question

part 1. A 9.00-m long string sustains a three-loop standing wave pattern as shown. The string has a mass of 45 g and under a tension of 50 N.

a. What is the frequency of vibration?

b. At the same frequency, you wish to see four loops, what tension you need to use.

Part 2.

a. Determine the shortest length of pipe, open at both ends, which will resonate at 256 Hz (so the first harmonics is 256Hz). The speed of sound is 343 m/s.

b. Determine the shortest length of pipe, open from one end and closed from the other end, which will resonate at 256 Hz (so the first harmonics is 256Hz). The speed of sound is 343 m/s.

c- For the tube in part b) calculate the third overtone

Please answer both parts and show work.

Answer 1

Part 1 :

Velocity of the wave in the string = √[T×L/m]

here, T = tension = 50 N

L = length of the string= 9 m

m = mass of the string = 45 g = 0.045 kg

Now, velocity of the wave= √[50×9/0.045] = 100 m/s

a)We know,

v/f =

here, f = frequency of the wave.

v = velocity of the wave = 100 m/s

   = wavelength = (2/3)×9 = 6 m

So, 100/f = 6

=> f = 100/6

=> f = 16.667 Hz

The frequency of the wave is 16.667 Hz

b) for 4 loops the wavelength will be = (2/4)×9 = 4.5 m

The velocity of the wave will be = 4.5×16.667 = 75 m/s

Let, the tension be T' N.

Then, √[T' × 9/0.045] = 75

=> 9×T'/0.045 = 5625

=> T' = 28.125 N

The new tension will be 28.125 N

Part 2 :

a)The shortest length of pipe will be=(v/2f) = (343/2×256) = 0.667 m

b) The shortest length of pipe will be = (v/4f)= (343/4×256)= 0.335 m

c) Third overtone = 7th harmonic.

The frequency of the 7th harmonic = 7×256 = 1792 Hz

The wavelength of the 3rd overtone =(343/1792) = 0.1914 m

Please comment if you have any doubt and like if it helps.