A standing wave pattern is created on a string with mass density = 3.2 × 10 -4 k

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 = 63 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 = 0.58 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 = 63 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 = 63 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*0.58*63/3= 24.36m/s
(1) =v/f = 24.36/63= 0.386[m]
(2) v= 24.36[m/s]
(3) v=(T/) -> T=v^2*
T= 24.36^2*3.2*10^-4= 0.189 N
(4) F=ma -> m=F/a , as a=9.8m/s^2
m= 0.189 / 9.8= 0.0193= 1.93*10^-2[kg]

(6) v=2Lf/n , n=2
v= 2*0.58*63/2= 36.54m/s
(5) =v/f = 36.54/63= 0.58[m]
(7) T=v^2* = 36.54^2*3.2*10^-4= 0.427 N
(8) m=F/a = 0.427/9.8= 0.0435= 4.35*10^-2 [kg]

(9) n=1: for fundamental frequency
v=2Lf/n = 2*0.58*63/1= 73.08m/s
T=v^2* = 73.08^2*3.2*10^-4= 1.709N
m=F/a = 1.709/9.8= 0.174= 1.743*10^-1[kg]

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