When listening to tuning forks of frequency 256 Hz and 260 Hz, one hears the following number of beats per second. (A) 0 (B) 2 (
1 answer:
Answer:
(C) 4 beats per second.
Explanation:
As we know that the no of beats can be calculated as.
No. of beats is equal to difference in the tuning forks frequencies.
So,
.
Substitute the values of frequencies of 2 tuning forks in the above equation.
Therefore the number of beats per second will be hear by the observer is 4 beats per second.
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To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.
In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.
By definition we know that the tensile strength is defined as
Where,
Tensile strength
F = Tensile Force
A = Cross-sectional Area
In the other hand we have that the shear strength is defined as
where,
Shear strength
Shear Force
Parallel Area
PART A) Replacing with our values in the equation of tensile strenght, then
Resolving for F,
PART B) We need here to apply the shear strength equation, then
In such a way that the material is more resistant to tensile strength than shear force.
