A standing wave pattern is created on a string with mass density = 3.2 × 10 -4 k
ID: 1473954 • Letter: A
Question
A standing wave pattern is created on a string with mass density = 3.2 × 10-4 kg/m. A wave generator with frequency f = 61 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 = 0.71 m. Initially the 3rd harmonic wave pattern is formed.
1)
What is the wavelength of the wave?
m
2)
What is the speed of the wave?
m/s
3)
What is the tension in the string?
N
4)
What is the mass hanging on the end of the string?
kg
5)
Now the hanging mass is adjusted to create the 2nd harmonic. The frequency is held fixed at f = 61 Hz.
What is the wavelength of the wave?
m
6)
What is the speed of the wave?
m/s
7)
What is the tension in the string?
N
8)
What is the mass hanging on the end of the string?
kg
9)
Keeping the frequency fixed at f = 61 Hz, what is the maximum mass that can be used to still create a coherent standing wave pattern?
kg
Explanation / Answer
f=nv/2L -> v=2Lf/n , n=3
v= 2*0.71*61/3= 28.87m/s
(1) =v/f = 28.87/61= 0.47[m]
(2) v= 28.87[m/s]
(3) v=(T/) -> T=v^2*
T= 28.87^2*3.2*10^-4= 0.2667 = 0.27[N] approx.
(4) F=ma -> m=F/a , as a=9.8m/s^2
m= 0.27/9.8= 0.02755= 2.76*10^-2[kg]
(6) v=2Lf/n , n=2
v= 2*0.71*61/2= 43.31m/s
(5) =v/f =43.31/61= 0.71[m]
(7) T=v^2* = 43.31^2*0.00032= 0.600 = 0.60[N]
(8) m=F/a = 0.6002/9.8= 0.0612= 6.12*10^-2[kg]
(9) n=1: for fundamental frequency
v=2Lf/n = 2*0.71*61/1= 86.62m/s
T=v^2* = 86.62^2*0.00032= 2.4N
m=F/a = 2.4/9.8= .245= 2.45*10^-1[kg]
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