Answer:
[tex]v_1 = 301 Hz[/tex]
[tex]v_2 = 601 \ \ Hz[/tex]
[tex]v_3 = 901 \ Hz[/tex]
Explanation:
From the question we are told that
The length of the string is [tex]l = 90 \ cm = 0.9 \ m[/tex]
The mass of the string is [tex]m_s = 3.5 \ g =0.0035 \ kg[/tex]
The distance from the bridge to the support post [tex]L = 62 \ c m = 0.62 \ m[/tex]
The tension is [tex]T = 540 \ N[/tex]
Generally the frequency is mathematically represented as
[tex]v = \frac{n}{2 * L } [\sqrt{ \frac{T}{\mu} } ][/tex]
Where n is and integer that defines that overtones
i.e n = 1 is for fundamental frequency
n = 2 first overtone
n =3 second overtone
Also [tex]\mu[/tex] is the linear density of the string which is mathematically represented as
[tex]\mu = \frac{m_s}{l}[/tex]
=> [tex]\mu = \frac{0.0035 }{ 0.9 }[/tex]
=> [tex]\mu = 0.003889 \ kg/m[/tex]
So for n = 1
[tex]v_1 = \frac{1}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ][/tex]
[tex]v_1 = 301 \ Hz[/tex]
So for n = 2
[tex]v_2 = \frac{2}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ][/tex]
[tex]v_2 = 601 \ Hz[/tex]
So for n = 3
[tex]v_3 = \frac{3}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ][/tex]
[tex]v =901 \ Hz[/tex]
