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calculate percentage error in determination of time period of a pendulum T=2π√l÷g where l and g are measured with ±1% and ±2% error.

calculate percentage error in determination of time period of a pendulum T=2π√l÷g where l and g are measured with ±1% and ±2% error.

Grade:11

1 Answers

Vikas TU
14149 Points
6 years ago
As we know that,
The equation of simple pendulum:
T=2π√L/g
From this equation, g will be as follows
⇒g=4π^2⋅L/T2
Now we take logarihm on both sides
lng=ln(4π^2)+lnL−2lnT
Taking differential
Δg/g=ΔL/L−2ΔT/T....(1)
Now for calculation of maximum error of g every error term should be taken positive.
Hence equation (1) becomes
Δg/g=ΔL/L+2ΔT/T
In percent form it becomes
Δg/g×100=ΔL/L×100+2ΔT/T×100..(2)
Now it is given that in measuring L=20cm the minimum length measurable by scale being 1mm or 0.1cm the error ΔL=0.1cm
∴ΔLL=0.120
Again the resolution of watch used for the measurement of T being 1s (the smallest graduation on the face of the watch).error for the measurement of time for 100 oscillations will be 1s i.e.ΔT=1100s=0.01s
Now time measured for 100 oscillations is 90s.So time period T=90100=0.9s
Hence ΔTT=0.010.9
Now the percent error in measurement of g by equation(2)
Δgg×100=ΔLL×100+2ΔTT×100
⇒Δgg×100=0.120×100+2×0.010.9×100
≈(0.5+2.2)%=2.7% ≈ 3%.

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