Question

The length of a simple pendulum is about 50 cm and is known to 1 mm accuracy. The time period of oscillations is about 1.0 s. The time of 100 oscillations is measured with a wrist watch of 1 s resolution. The percentage error in determination of g is



​

Answer 1

Answer:

5%

Here, T=2π

g

L

Squaring both sides, we get, T

2

=

g

4π

2

L

org=

T

2

4π

2

L

The relative error in g is,

g

Δg

=

L

ΔL

+2

T

ΔT

Here, T=

n

t

and ΔT=

n

Δt

∴

T

ΔT

=

t

Δt

The errors in both L and t are the least count errors.

∴

g

Δg

=

10

0.1

+2(

50

1

)=0.01+0.04=0.05

The percentage error in g is

g

Δg

×100=

L

ΔL

×100+2(

T

ΔT

)×100=[

L

ΔL

+2(

T

ΔT

)]×100=0.05×100=5%

Answer 2

Given : The length of a simple pendulum is about 50 cm and is known to 1 mm accuracy. The time period of oscillations is about 1.0 s. The time of 100 oscillations is measured with a wrist watch of 1 s resolution.

To Find : The percentage error in determination of g is

Explanation:

$t = 2\pi \sqrt{ \frac{l}{g} }$

squaring both sides

${t}^{2} = 4 {\pi}^{2} \frac{l}{g}$

$g = 4 {\pi}^{2} \frac{l}{ {t}^{2} }$

∆g/g = ∆l/l + 2∆t/t

100 × ∆g/g = 100 × ∆l/l + 100 × 2∆t/t

∆l/l = 1 / (50 × 10) = 1/500

∆t/t = 1/(100 × 1) = 1/100

100 × ∆g/g = 100 × 1/500+ 100 × 2 × 1/100

100 × ∆g/g = 0.2 + 2

100 × ∆g/g = 2.2

The percentage error in determination of g is 2.2 .

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