Question #47b8a

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Question #47b8a

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Answer 1 · 2017-12-26T16:22:26

See the explanation below

Explanation:

The length of the pendulum is $= l$ $=l$

The period of the pendulum is $= T$ $=T$

The period of a simple pendulum is

$T = 2 π \sqrt{\frac{l}{g}}$ $T=2pisqrt(l/g)$

Taking natural logarithms on both sides

$ln T = ln (2 π) + \frac{1}{2} ln (l) - \frac{1}{2} ln (g)$ $lnT=ln(2pi)+1/2ln(l)-1/2ln(g)$

Differentiating the equation

$\frac{Δ T}{T} = 0 + \frac{1}{2} \frac{Δ l}{l} - \frac{1}{2} \frac{Δ g}{g}$ $(DeltaT)/T=0+1/2(Deltal)/l-1/2(Deltag)/g$

You cannot substract errors

Therefore,

$∣ ∣ ∣ \frac{Δ T}{T} ∣ ∣ ∣ = \frac{1}{2} ∣ ∣ ∣ \frac{Δ l}{l} ∣ ∣ ∣ + \frac{1}{2} ∣ ∣ ∣ \frac{Δ g}{g} ∣ ∣ ∣$ $|(DeltaT)/T|=1/2|(Deltal)/l|+1/2|(Deltag)/g|$

The error in measuring the length is $\frac{Δ l}{l} = 2 %$ $(Deltal)/l=2%$

The error in measuring the acceleration due to gravity is $\frac{Δ g}{g} = 2 %$ $(Deltag)/g=2%$

So,

$\frac{Δ T}{T} = \frac{1}{2} \cdot 2 % + \frac{1}{2} \cdot 2 % = 2 %$ $(DeltaT)/T=1/2*2%+1/2*2%=2%$

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Question #47b8a

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