Answer:
Part 1:
The frequency of a vibrating string in its fundamental mode is given by:
f = (1/2L) √(T/μ)
where L is the length of the string, T is the tension in the string, and μ is the linear mass density (mass per unit length) of the string.
In this problem, f = 303 1/s, L = 42.7 cm = 0.427 m, and μ = m/L, where m is the mass of the vibrating segment. Substituting these values into the formula, we get:
303 1/s = (1/2 × 0.427 m) √(T/(1.04 g/0.427 m))
303 1/s = (1/2 × 0.427 m) √(T/0.00243 kg/m)
303 1/s = 0.0949 √T
T = (303 1/s / 0.0949)^2 × 0.00243 kg/m
T = 4.29 N
Therefore, the tension in the string is 4.29 N.
Part 2:
When a string vibrates in two segments, it is vibrating in its second harmonic or first overtone, which has two segments of equal length vibrating in opposite directions. The frequency of the second harmonic is given by:
f = (1/L) √(T/μ) × 2
where L, T, and μ have the same meaning as in Part 1. Substituting the values we found in Part 1, we get:
f = (1/0.427 m) √(4.29 N / 0.00243 kg/m) × 2
f = 712.7 1/s
Therefore, the frequency of the string when it vibrates in two segments is 712.7 1/s.