Answer:
11.07Hz
Explanation:
Check the attachment for diagram of the standing wave in question.
Formula for calculating the fundamental frequency Fo in strings is V/2L where;
V is the velocity of the wave in string
L is the length of the string which is expressed as a function of its wavelength.
The wavelength of the string given is 1.5λ(one loop is equivalent to 0.5 wavelength)
Therefore L = 1.5λ
If L = 3.0m
1.5λ = 3.0m
λ = 3/1.5
λ = 2m
Also;
V = √T/m where;
T is the tension = 0.98N
m is the mass per unit length = 2.0g = 0.002kg
V = √0.98/0.002
V = √490
V = 22.14m/s
Fo = V/2L (for string)
Fo = 22.14/2(3)
Fo = 22.14/6
Fo = 3.69Hz
Harmonics are multiple integrals of the fundamental frequency. The string in question resonates in 2nd harmonics F2 = 3Fo
Frequency of the wave = 3×3.69
Frequency of the wave = 11.07Hz
Answer:
550000000N/m
Explanation:
Given that a copper wire has a radius of 2.9 mm. When forces of a certain equal magnitude but opposite directions are applied to the ends of the wire, the wire stretches by 5.0×10−3 of its original length.
Original length L = 0.005L
the strain = extension/ original length
the strain = 0.005L / L
the strain = 0.005
Young modulus = stress / strain
11 × 10^10 = stress / 0.005
Cross multiply
Stress = 11 × 10^10 × 0.005
Stress = 550 000000 N/m
Therefore, the tensile stress on the wire is 550000000 N/m.