Part a) In this part, the oscillator was configured such that nodes are formed a
ID: 1526274 • Letter: P
Question
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?
2.0Explanation / 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