Question

The time period of oscillations of a simple pendulum is $T=2\pi \sqrt{\dfrac{L}{g}}$. L is about 10 cm and is known to 1 mm accuracy. The time period of oscillations is about 0.5 s. The time of 100 oscillations is measured with a wristwatch of 1 s resolution. What is the accuracy in the determination of ‘g’?
A. 1%
B. 2%
C. 5%
D. 7%

Answer 1

As given in the question statement, the formula for calculating the time period of oscillations should be used to solve this problem. Firstly, we will derive the expression of the percentage error in the ‘g’ in terms of the percentage error in the ‘T’ and the percentage error in the ‘L’. Then, we will substitute the given values in the expression obtained to find the value of the accuracy in the determination of ‘g’.

Formula used:
$T=2\pi \sqrt{\dfrac{L}{g}}$

Complete step-by-step answer:
The formula for calculating the time period of oscillations of a simple pendulum is given as follows.
$T=2\pi \sqrt{\dfrac{L}{g}}$
Where L is the length of the string and g is the gravitational constant.
Now, we will derive the expression for the percentage error in the terms involved in the above equation.
The percentage error in the time period of oscillations of a simple pendulum is,
$\dfrac{\Delta T}{T}\times 100=\dfrac{\Delta t}{t}\times 100$…… (1)
The percentage error in the length of the string of a simple pendulum is,
$\dfrac{\Delta L}{L}\times 100$…… (2)
The time period of oscillations of a simple pendulum in terms of the gravitational constant is,

& T=2\pi \sqrt{\dfrac{L}{g}} \\\ & \Rightarrow g=4{{\pi }^{2}}\dfrac{L}{{{T}^{2}}} \\\ \end{aligned}$$ Therefore, the percentage error in the gravitational constant is obtained by combining the equations (1) and (2). $$\dfrac{\Delta g}{g}\times 100=\dfrac{\Delta L}{L}\times 100+2\left( \dfrac{\Delta T}{T}\times 100 \right)$$….. (3) Firstly, we will find the percentage error in the time period of oscillations of a simple pendulum $$\dfrac{\Delta T}{T}\times 100=\dfrac{\Delta t}{t}\times 100$$ Substitute the given values in the above equation. $ \dfrac{\Delta T}{T}\times 100=\dfrac{1}{0.5\times 100}\times 100 $ $\Rightarrow \dfrac{\Delta T}{T}\times 100=2\% $ Therefore, the percentage error in the time period of oscillations of a simple pendulum is 2% Now we will find the percentage error in the length of the string of a simple pendulum $ \dfrac{\Delta L}{L}\times 100=\dfrac{0.1}{10}\times 100 $ $ \Rightarrow \dfrac{\Delta L}{L}\times 100=1\% $ Therefore, the percentage error in the length of the string of a simple pendulum is 1% Substitute these obtained values of the percentage error in the equation (3) $\dfrac{\Delta g}{g}\times 100=1\%+2\left( 2\% \right) $ $ \Rightarrow \dfrac{\Delta g}{g}\times 100=1\%+4\% $ $ \dfrac{\Delta g}{g}\times 100=5\% $ As the value of the accuracy in determination of ‘g’ obtained is equal to 5%, thus, the option (C) is correct. **So, the correct answer is “Option (C)”.** **Note:** There is a bit of confusion in obtaining the formula for calculating the percentage error in ‘g’. Because the percentage error in the length of the string of a simple pendulum is considered as it is because its power is 1 when we determined the expression of the time period of oscillations in terms of ‘g’ and the percentage error in the time period of the string of a simple pendulum is considered as double its value because its power is 2 when we determined the expression of the time period of oscillations in terms of ‘g’.