(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%.
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Answer:
Angular momentum = 0.7 kg.m²/s
Angular velocity = 583.3 rad/s
Explanation:
1. The torque τ is related to the angular momentum L by the relation
τ = ΔL/Δt
ΔL = τΔt
τ = 10 N. m
Δt = 70 ms = 70 × 10⁻³s
ΔL = (10 N. m) × (70 × 10⁻³s) = 700 × 10⁻³ kg.m²/s = 0.7 kg.m²/s
2. The rotational inertia I relates the angular momentum L to the angular velocity w
L = Iw
w = L/I
L = 0.7 kg.m²/s
I = 1.2 × 10⁻³ kg.m²
w = (0.7 kg.m²/s)/(1.2 × 10⁻³ kg.m²) = 583.3 rad/s
With the values you've given, only velocity can be found.
Acceleration is rate of change of velocity
d= 250s
t= 17s
a= d/t
=

= 4.7
You would need to use the equation a= (v-u)÷t
You need to substitute in the correct numbers.
a= (10-20)÷1
Your answer is -10m/s^2