Answer:
The linear mass density is of the string [tex]\mu= 5.51*10^{-3} kg / m[/tex]
Explanation:
From the question we are told that
The distance between wall and pulley is [tex]d = 0.405m[/tex]
The mass on the hook is [tex]m = 25.4\ kg[/tex]
The frequency of oscillation is [tex]f = 261.6 Hz[/tex]
Generally, the frequency of oscillation is mathematically represented as
[tex]f = \frac{1}{2d} \sqrt{\frac{T}{\mu} }[/tex]
Where T is the tension mathematically represented as
T = mg
Substituting values
[tex]T = 25.4 *9.8[/tex]
[tex]=248.92N[/tex]
[tex]\mu[/tex] is the mass linear density
Making [tex]\mu[/tex] the subject of the formula above
[tex]\mu = \frac{T}{(2df)^2}[/tex]
Substituting values
[tex]\mu = \frac{248.92}{(2 * 0.405 * 261.6)^2}[/tex]
[tex]\mu= 5.51*10^{-3} kg / m[/tex]
Answer:
0.005550 Kg/m
Explanation:
The picture attached below shows the full explanation
