Question

The period of oscillation of a simple pendulum is T=2π√L/g. measured value of L is 10cm CM known to 1 mm accuracy and time for 100 oscillations of the pendulum is found using the wrist watch of 1s resolution.The periodic oscillation is about 0.5s. what is the accuracy in the determination of g?

Answer is 5%​

Answer 1

If picture is not clear...

Than

T =2π√( L/g )

t/n = 2π√L/g

t= 4π²Ln²/g

g = 4π²L/t²

here n is no of oscillations

so,

∆g/g = ∆L/L + 2∆t/t

given ,

L = 100 cm

∆L = 1mm = 0.1 cm

t = 200 sec

∆t =0.1 sec

now,

∆g/g = 0.1/100 + 2× 0.1/200

percentage error % of g = ∆g/g ×100

=0.1/100 ×100 + 0.1×100/100

=0.1 + 0.1= 0.2 %

Answer 2

Given :

$T=2\pi\sqrt{\dfrac{l}{g}}\\\\\implies T^2=4\pi^2\times \dfrac{l}{g}\\\\\implies g=\dfrac{l\times 4pi^2}{T^2}$

We have to calculate the error in determination in the value of g . So we must cancel all the constants that is π should be removed from our calculations because constants never cause an error .

Now students mostly make mistake in this part that they subtract errors . Errors will always be added and never subtracted !

$\dfrac{\delta g}{g}=\dfrac{\delta l}{l}+2\dfrac{\delta t}{t}$

NOTE :

Powers should be multiplied like I have done . When you are dividing or multiplying always add the error .

l = 10 cm [ given ]

δl = 1 mm [ given ]

= > δl = 0.1 cm

t = 0.5 s

δt = 1/100 [ for 100 oscillations ]

= > δt = 0.01 s

$\dfrac{\delta g}{g}=\dfrac{\delta l}{l}+2\dfrac{\delta t}{t}\\\\\implies \dfrac{\delta g}{g}=\dfrac{0.1}{10}+2(\dfrac{0.01}{0.5})\\\\\implies \dfrac{\delta g}{g}=0.01+2\times 0.02\\\\\implies \dfrac{\delta g}{g}=0.01+0.04\\\\\implies \dfrac{\delta g}{g}=0.05$

$\textsf{Accuracy percentage will be :}\\\\\dfrac{\delta g}{g}\times 100\%\\\\\implies 0.05 \times 100\%\\\\\implies \boxed{5\%}$