(a) The tension the musician must stretch it is 147.82 N.
(b) The percent increase in tension is needed to increase the frequency is 26%.
<h3>Tension in the string</h3>
v = √T/μ
where;
- v is speed of the wave
- T is tension
- μ is mass per unit length = 0.0144 kg / 0.6 m = 0.024 kg/m
v = Fλ
in fundamental mode, v = F(2L)
v = 2FL
v = 2 x 65.4 x 0.6 = 78.48 m/s
v = √T/μ
v² = T/μ
T = μv²
T = 0.024 x (78.48)²
T = 147.82 N
<h3>When the frequency is 73.4 Hz;</h3>
v = 2FL = 2 x 73.4 x 0.6 = 88.08 m/s
T = μv²
T = (0.02)(88.08)²
T = 186.19 N
<h3>Increase in the tension</h3>
= (186.19 - 147.82)/(147.82)
= 0.26
= 0.26 x 100%
= 26 %
Thus, the tension the musician must stretch it is 147.82 N.
The percent increase in tension is needed to increase the frequency is 26%.
Learn more about tension here: brainly.com/question/24994188
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