Answer:
a) Δf = 0.7 n
, e) f = (15.1 ± 0.7) 10³ Hz
Explanation:
This is an error about the uncertainty or error in the calculated quantities.
Let's work all the magnitudes is the SI system
The frequency of oscillation is
f = n / 2π L² √( E /ρ)
where n is an integer
Let's calculate the magnitude of the oscillation
f = n / 2π (0.2335)² √ (210 10⁹/7800)
f = n /0.34257 √ (26.923 10⁶)
f = n /0.34257 5.1887 10³
f = 15.1464 10³ n
a) We are asked for the uncertainty of the frequency (Df)
Δf = | df / dL | ΔL + df /dE ΔE + df /dρ Δρ
in this case no error is indicated in Young's modulus and density, so we will consider them exact
ΔE = Δρ = 0
Δf = df /dL ΔL
df = n / 2π √E /ρ | -2 / L³ | ΔL
df = n / 2π 5.1887 10³ | 2 / 0.2335³) 0.005 10⁻³
df = n 0.649
Absolute deviations must be given with a single significant figure
Δf = 0.7 n
b, c) The uncertainty with the width and thickness of the canteliver is associated with the density
In your expression there is no specific dependency so the uncertainty should be zero
The exact equation for the natural nodes is
f = n / 2π L² √ (E e /ρA)
where A is the area of the cantilever and its thickness,
In this case, they must perform the derivatives, calculate and approximate a significant figure
Δf = | df / dL | ΔL + df /de Δe + df /dA ΔA
Δf = 0.7 n + n 2π L² √(E/ρ A) | ½ 1/√e | Δe
+ n / 2π L² √(Ee /ρ) | 3/2 1√A23 |
the area is
A = b h
A = 24.9 3.3 10⁻⁶
A = 82.17 10⁻⁶ m²
DA = dA /db ΔB + dA /dh Δh
dA = h Δb + b Δh
dA = 3.3 10⁻³ 0.005 10⁻³ + 24.9 10⁻³ 0.005 10⁻³
dA = (3.3 + 24.9) 0.005 10⁻⁶
dA = 1.4 10⁻⁷ m²
let's calculate each term
A ’= n / 2π L² √a (E/ρ A) | ½ 1 /√ e | Δe
A ’= n/ 2π L² √ (E /ρ) | ½ 1 / (√e/√ A) |Δe
A ’= 15.1464 10³ n ½ 1 / [√ (24.9 10⁻³)/ √ (81.17 10⁻⁶)] 0.005 10⁻³
A '= 0.0266 n
A ’= 2.66 10⁻² n
A ’’ = n / 2π L² √ (E e /ρ) | 3/2 1 /√A³ |
A ’’ = n / 2π L² √(E /ρ) √ e | 3/2 1 /√ A³ | ΔA
A ’’ = n 15.1464 10³ 3/2 √ (24.9 10⁻³) /√ (82.17 10⁻⁶) 3 1.4 10⁻⁷
A ’’ = n 15.1464 1.5 1.5779 / 744.85 1.4 10⁴
A ’’ = 6,738 10²
we write the equation of uncertainty
Δf = n (0.649 + 2.66 10⁻² + 6.738 10²)
The uncertainty due to thickness is
Δf = 3 10⁻² n
The uncertainty regarding the area, note that this magnitude should be measured with much greater precision, specifically the height since the errors of the width are very small
Δf = 7 10² n
d) Δf = 7 10² n
e) the natural frequency n = 1
f = (15.1 ± 0.7) 10³ Hz
