Question 3: Standing wave You are asked to determine the frequency of a string o

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Question 3: Standing wave You are asked to determine the frequency of a string oscillator that was shown in class. A schematic diagram is shown here. The length of the string is 2.0 m. The weight of the 2.0 m long string is 0.003 kg. The velocity of a wave in a string is given by v=sqrt{T/mu} where T is the tension of the string and mu is the linear mass density (in kg/m, total mass divided by the total length) of the string (see the Figure below). Part a) In this part, the oscillator was configured such that nodes are formed at both ends of the string. When a mass of 0.1 kg was hung, a standing wave of 4 antinodes was observed. What is the frequency of this wave? Part b) The oscillator is now reconfigured so that a node will be formed at the pulley but an antinode will be formed at the oscillator end. The same mass of 0.1 kg was used and the frequency was adjusted such that a standing wave of 4 antinodes was observed. What is the frequency of this wave?

Question 3: Standing wave You are asked to determine the frequency of a string oscillator that was shown in clas A schematic diagram is shown here. The length of the string is 2.0 m. The weight of the 2.0 m long string is 0.003 kg. The velocity of a wave in a string is given by v VTTH where Tis the tension of the string and H is the linear mass density (in kg/m, total mass divided by the total length) of the string (see the Figure below). Oscillator Part a) n this part, the oscillator was configured such that nodes are formed at both ends of the string. When a mass of 0.1 kg was hung, a standing wave of 4 antinodes was observed. What is the frequency of this wave? Part b The oscillator is now reconfigured so that a node will be formed at the pulley but an antinode will be formed at the oscillator end. The same mass of 0.1 kg was used and the frequency was adjusted such that a standing wave of4 antinodes was observed. What is the frequency of this wave?

Explanation / Answer

from the given data

linear mass density of string, mue = 0.003/2

= 0.0015 kg/m

Tension in the string, T = m*g

= 0.1*9.8

= 0.098 N

a) no of loops, n = 4

speed of the wave on the string, v = sqrt(T/mue)

= sqrt(0.098/0.0015)

= 8.1 m/s

now, use, f = n*v/(2*L)

= 4*8.1/(2*2)

= 8.1 Hz

b)

no of loops, n = 3.5

now, use, f = n*v/(2*L)

= 3.5*8.1/(2*2)

= 7.1 Hz

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