Afterthought
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So I just did a lab on resonant frequency, and have to find the error of the period: $$T = 2π\sqrt{m/k}$$
m and k were measured with an error of Δm and Δk. My teacher didn't explain well how to do errors, so I just want to know if I did this right.
Prorogation of Error equations:
$$y=x^n ⇒ Δy=nx^{n-1}Δx$$ $$y=x_1/x_2 ⇒ Δy=y\sqrt{(Δx_1/x_1)^2+(Δx_2/x_2)^2}$$
Attempt at solution:
$$Δ(m/k)^{0.5}=0.5(m/k)^{-0.5}Δ(m/k)$$ $$Δ(m/k)=(m/k)\sqrt{(Δm/m)^2+(Δk/k)^2}$$ $$ Therefore, ΔT = 2π * 0.5(m/k)^{-0.5}(m/k)\sqrt{(Δm/m)^2+(Δk/k)^2}=π\sqrt{(Δm)^2/(mk)+(m(Δk)^2)/k^3}$$
m and k were measured with an error of Δm and Δk. My teacher didn't explain well how to do errors, so I just want to know if I did this right.
Prorogation of Error equations:
$$y=x^n ⇒ Δy=nx^{n-1}Δx$$ $$y=x_1/x_2 ⇒ Δy=y\sqrt{(Δx_1/x_1)^2+(Δx_2/x_2)^2}$$
Attempt at solution:
$$Δ(m/k)^{0.5}=0.5(m/k)^{-0.5}Δ(m/k)$$ $$Δ(m/k)=(m/k)\sqrt{(Δm/m)^2+(Δk/k)^2}$$ $$ Therefore, ΔT = 2π * 0.5(m/k)^{-0.5}(m/k)\sqrt{(Δm/m)^2+(Δk/k)^2}=π\sqrt{(Δm)^2/(mk)+(m(Δk)^2)/k^3}$$
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