Answer:
0.98 g/m
Explanation:
Note: Since Tension and frequency are constant,
Applying,
F₁²M₁ = F₂²M₂............... Equation 1
Where F₁ = Frequency of the G string, F₂ = Frequency of the A string, M₁ = mass density of the G string, M₂ = mass density of the A string.
make M₂ the subject of the equation
M₂ = F₁²M₁/F₂²............... Equation 2
From the question,
Given: F₁ = 196 Hz, M₁ = 0.31 g/m, F₂ = 110 Hz
Substitute these values into equation 2
M₂ = 196²(0.31)/110²
M₂ = 0.98 g/m
To solve this exercise it is necessary to use the concepts related to Difference in Phase.
The Difference in phase is given by
Where
Horizontal distance between two points
Wavelength
From our values we have,
The horizontal distance between this two points would be given for
Therefore using the equation we have
Therefore the correct answer is C.
If it attractive it has opposite pole and if it repulsive it has same pole
