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Question: The time period of oscillation of simple pendulum given by \(T = 2\pi \sqrt {\dfrac{L}{g}} \) where\...

The time period of oscillation of simple pendulum given by T=2πLgT = 2\pi \sqrt {\dfrac{L}{g}} whereL=(200±0⋅1)cmL = \left( {200 \pm 0 \cdot 1} \right)cm. The time period, T=4sT = 4s and the time of 100 oscillations is measured using a stopwatch of least count 0.1s. the percentage error in g is:
A) 0.1%.
B) 1.5%.
C) 2%.
D) 4%.

Explanation

Solution

Oscillations is defined as the periodic variation between two values about central values. The time period is defined as the time taken by the wave to cross one wavelength. The time period of the oscillation depends upon the acceleration due to gravity and length of the pendulum.

Step by step solution:
It is given in the problem that time period of oscillation of simple pendulum given by T=2πLgT = 2\pi \sqrt {\dfrac{L}{g}} whereL=(200±0⋅1)cmL = \left( {200 \pm 0 \cdot 1} \right)cm the time period, T=4sT = 4s and the time of 100 oscillations is measured using a stopwatch of least count 0.1s and we need to find the percentage error in the value of g.
The time period of the pendulum is given by,
⇒T=2πLg\Rightarrow T = 2\pi \sqrt {\dfrac{L}{g}}
⇒T=2πLg\Rightarrow T = 2\pi \sqrt {\dfrac{L}{g}}
⇒T2=(2πLg)2\Rightarrow {T^2} = {\left( {2\pi \sqrt {\dfrac{L}{g}} } \right)^2}
⇒T2=4π2×Lg\Rightarrow {T^2} = 4{\pi ^2} \times \dfrac{L}{g}
⇒g=4π2×LT2\Rightarrow g = 4{\pi ^2} \times \dfrac{L}{{{T^2}}}
⇒Δgg=(ΔLL+2ΔTT)×100\Rightarrow \dfrac{{\Delta g}}{g} = \left( {\dfrac{{\Delta L}}{L} + 2\dfrac{{\Delta T}}{T}} \right) \times 100
Length of pendulum is 2 m and the error in the length is 0.1 cm also the time period is 400 sec. for 100 oscillations and least count is 0.1sec.
⇒Δgg=(ΔLL+2ΔTT)×100\Rightarrow \dfrac{{\Delta g}}{g} = \left( {\dfrac{{\Delta L}}{L} + 2\dfrac{{\Delta T}}{T}} \right) \times 100
⇒Δgg=(0⋅0012+2×0⋅1400)×100\Rightarrow \dfrac{{\Delta g}}{g} = \left( {\dfrac{{0 \cdot 001}}{2} + 2 \times \dfrac{{0 \cdot 1}}{{400}}} \right) \times 100
⇒Δgg=(0⋅0005+0⋅0005)×100\Rightarrow \dfrac{{\Delta g}}{g} = \left( {0 \cdot 0005 + 0 \cdot 0005} \right) \times 100
⇒Δgg=(0⋅001)×100\Rightarrow \dfrac{{\Delta g}}{g} = \left( {0 \cdot 001} \right) \times 100
⇒Δgg=0⋅1%\Rightarrow \dfrac{{\Delta g}}{g} = 0 \cdot 1\%.
The percentage error of the acceleration due to gravity is equal toΔgg=0⋅1%\dfrac{{\Delta g}}{g} = 0 \cdot 1\% .

The correct answer for this problem is option A.

Note: It is advisable to students to understand and remember the formula of the time period of the pendulum and the formula for taking out the error in the time period. There will be error in the time period if there is error in the measurement of the length of the pendulum and if there is error in the value of acceleration due to gravity.