Answer:
The value of g is [tex]g =76.2 m/s^2[/tex]
Explanation:
From the question we are told that
The mass of the weight is [tex]m = 1.30 kg[/tex]
The spring constant [tex]k = 1.73 g/m = 1.73 *10^{-3} \ kg/m[/tex]
The second harmonic frequency is [tex]f = 100 \ Hz[/tex]
The number of oscillation is [tex]N = 200[/tex]
The time taken is [tex]t = 315 \ s[/tex]
Generally the frequency is mathematically represented as
[tex]f = \frac{v}{\lambda}[/tex]
At second harmonic frequency the length of the string vibrating is equal to the wavelength of the wave generated
[tex]l = \lambda[/tex]
Noe from the question the vibrating string is just half of the length of the main string so
Let assume the length of the main string is [tex]L[/tex]
So [tex]l = \frac{L}{2}[/tex]
The velocity of the vibrating string is mathematically represented as
[tex]v = \sqrt{\frac{T}{\mu} }[/tex]
Where T is the tension on the string which can be mathematically represented as
[tex]T = mg[/tex]
So
[tex]v = \sqrt{\frac{mg}{k} }[/tex]
Then
[tex]f = \frac{v}{\frac{L}{2} }[/tex]
=> [tex]v = \frac{fL }{2}[/tex]
=> [tex]\sqrt{\frac{mg}{k} } = \frac{fL}{2}[/tex]
=> [tex]g = \frac{f^2 L^2 \mu}{4m}[/tex]
substituting values
[tex]g = \frac{(100) * (1.73 *10^{-3} )}{(4 * 1.30)} L^2[/tex]
[tex]g = 3.326 m^{-1} s^{-2} L^2[/tex]
Generally the period of oscillation is mathematically represented as
[tex]T_p = 2 \pi \sqrt{\frac{L}{g} }[/tex]
=> [tex]L = \frac{T^2 g}{4 \pi ^2}[/tex]
The period can be mathematically evaluated as
[tex]T_p = \frac{t}{N}[/tex]
substituting values
[tex]T_p = \frac{315}{200}[/tex]
[tex]T_p = 1.575 \ s[/tex]
Therefore
[tex]L = \frac{1.575^2 * g }{4 \pi ^2}[/tex]
[tex]L = 0.0628 ^2 g[/tex]
so
[tex]g = 3.326 m^{-1} s^{-2} L^2[/tex]
substituting for L
[tex]g = 3.326 ((0.0628) g)^2[/tex]
=> [tex]g = \frac{1}{(3.326)* (0.0628)^2}[/tex]
[tex]g =76.2 m/s^2[/tex]
