1) 83.6 N
2) 1434 Hz
Explanation:
1)
The fundamental frequency of the standing waves on a string is given by the equation
[tex]f_1=\frac{1}{2L}\sqrt{\frac{T}{\mu}}[/tex]
where:
L is the length of the string
T is the tension in the string
[tex]\mu[/tex] is the mass per unit length, which can be rewritten as
[tex]\mu=\frac{m}{L}[/tex], where m is the mass of the string
For the string in this problem, we have:
L = 76.2 cm = 0.762 m is the length of the string
[tex]m=0.48 g = 0.48\cdot 10^{-3} kg[/tex] is the mass
[tex]f_1=239 Hz[/tex] is the fundamental frequency
Re-arranging the equation for T, we find the tension in the string:
[tex]T=(2Lf_1)^2 \mu = (2Lf_1)^2 \frac{m}{L}=(2(0.762)(239))^2 \frac{0.48\cdot 10^{-3}}{0.762}=83.6 N[/tex]
2)
In this problem, we see that the string vibrates in six segments.
A standing wave on a string has different modes of vibrations, characterized by different frequencies:
- In the fundamental mode ([tex]f_1[/tex]), the string vibrates in one segment
- In the second harmonic ([tex]f_2[/tex]), the string vibrates in 2 segments
... and so on
The frequency of the nth-harmonic is given by
[tex]f_n = nf_1[/tex]
where [tex]f_1[/tex] is the fundamental frequency.
In this problem, we know that
[tex]f_1=239 Hz[/tex]
And the string is vibrating in 6 modes, so we are finding the frequency of the 6th harmonic, which is
[tex]f_6 = 6f_1 = 6(239)=1434 Hz[/tex]
