Question 116: Oscillations

Question

The period of oscillation of a simple pendulum is $$T = 2\pi \sqrt {{L \over g}} $$. Measured value of 'L' is 1.0 m from meter scale having a minimum division of 1 mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01 s resolution. The percentage error in the determination of 'g' will be :

Accepted Answer

Given, $$T = 2\pi \sqrt {{L \over g}} $$ .... (i)
where, time period, T = 1.95 s
Length of string, l = 1 m
Acceleration due to gravity = g
Error in time period, $$\Delta$$T = 0.01 s = 10
$$-$$2
s
Error in length, $$\Delta$$L = 1 mm = 1 $$	imes$$ 10
$$-$$3
m
Squaring Eq. (i) on both sides, we get
$${T^2} = 4{\pi ^2}{L \over g}$$
$$ \Rightarrow g = 4{\pi ^2}{L \over {{T^2}}}$$
$$ \Rightarrow {{\Delta g} \over g} = {{\Delta L} \over L} + {{2\Delta T} \over T} = {{{{10}^{ - 3}}} \over 1} + {{2 	imes {{10}^{ - 2}}} \over {1.95}}$$
$$ = {10^{ - 3}} + 1.025 	imes {10^{ - 2}}$$
$$ = {10^{ - 3}} + 10.25 	imes {10^{ - 3}}$$
$$ = 11.25 	imes {10^{ - 3}}$$
$$\because$$ $$\Delta g/g 	imes 100 = 11.25 	imes {10^{ - 3}} 	imes {10^2}$$
$$ = 1.125\%  \simeq 1.13\% $$

Answer

1.30%

Answer

1.33%

Answer

1.13%

Answer

1.03%

Oscillations

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