A standing wave pattern is created on a string with mass density ? = 3.2 Times 1

Question

A standing wave pattern is created on a string with mass density ? = 3.2 Times 10-4 kg/m. A wave generator with frequency f = 62 Hz is attached to one end of the string and the other end goes over a pulley and is connected to a mass (ignore the weight of the string between the pulley and mass). The distance between the generator and pulley is L = 0.55 m. Initially the 3rd harmonic wave pattern is formed. What is the wavelength of the wave? m What is the speed of the wave? m/s What is the tension in the string? N What is the mass hanging on the end of the string? kg mass is adjusted to create the 2nd harmonic. The frequency is held fixed at f = 62 Hz. What is the wavelength of the wave? m What is the speed of the wave? m/s What is the tension in the string? N What is the mass hanging on the end of the string? kg Keeping the frequency fixed at f = 62 Hz, what is the maximum mass that can be used to still create a coherent standing wave pattern?

Explanation / Answer

1) lamda = (2/3)*L

= (2/3)*0.55
= 0.367 m

2)

v = lamda*f = 0.367*62 = 22.73 m/s

3)

v = sqrt(T/mue)

T = v^2*mue = 22.73^2*3.2*10^-4 = 0.1654 N

4)

m = T/g = 0.1653/9.8 = 0.01687 kg = 16.87 grams

5)

lamda = L = 0.55 m

6)

v = lamda*f = 0.55*62 = 34.1 m/s

7)
v = sqrt(T/mue)

T = v^2*mue = 34.1^2*3.2*10^-4 = 0.3721 N

8)
m = T/g = 0.3721/9.8 = 0.0378 kg = 37.8 grams


9)

let m is the maximum mass

f = sqrt(T/mue)/(2*L)

f = sqrt(m*g/mue)/(2*L)

f*2*L = sqrt(m*g/mue)

m = 4*f^2*L^2*mue/g

= 4*62^2*0.55^2*3.2*10^-4/9.8

= 0.1518 kg

= 151.8 grams

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