Answer to Question #175841 in Mechanics | Relativity for Satyabrat

Question

The length (𝑙) and Time Period (𝑇) of a simple pendulum are related as: 𝑇 = 2𝜋√

𝑙

𝑔

, where 

𝑔 is acceleration due to gravity. In an experiment the following values for 𝑙 and 𝑇 are found:

𝑙 (in cm) 𝑇 (in seconds) 𝑙 (in cm) 𝑇 (in seconds) 𝑙 (in cm) 𝑇 (in seconds)

50 1.41 70 1.67 90 1.90

55 1.48 75 1.73 95 1.95

60 1.55 80 1.79 100 2.00

65 1.61 85 1.84

You can verify, if you like, taking 𝑔 ≈ 981 𝑐𝑚 𝑠

−2

. The values of 𝑇 are approximated to the 

second place after decimal. Draw the 𝑙 − 𝑇 graph (do it manually, don’t use a computer

Answer 1

Explanations & Calculations

• This experiment is done to verify the value of gravitational acceleration: g & the data is usually plotted as $\small T^2 \, \text{vs}\, L$.
• And By the gradient of the graph, the value of g can be calculated.
• To make the hand drawing of the plot easier, the values of $\small T^2$ can be written in whole numbers & powers as done here.

• Refer to the figures attached.

• Table of data

• Plot

• Omit any extraneous values that may be some mistakes (such as L:80cm & T:1.70s) & plot the graph.

• To calculate the gradient of the graph, try to find farthest 2 points that lie on the best fit line & calculate the tangent value.

$\qquad\qquad \begin{aligned} \small m&=\small \tan \theta =\frac{(400-219)\times10^{-2 }}{1-0.55}\\ &=\small 402.22\times 10^{-2} \end{aligned}$

• Then compare the values,

$\qquad\qquad \begin{aligned} \small T&=\small 2\pi \sqrt{\frac{L+\epsilon}{g}}\\ \small T^2 &=\small 4\pi^2 \frac{(L+\epsilon)}{g}\\ \small T^2 &=\small \frac{4\pi^2}{g}\cdot L+\frac{4\pi^2 }{g}\cdot\epsilon\\ &\downarrow\\ \small y&=\small mx+c\\ \\ \small m&=\small 4.0222=\frac{4\pi^2}{g}\\ \small g&=\small \frac{4\pi^2}{4.0222}=9.815\\ &\approx\small \bold{9.82ms^{-2}} \end{aligned}$ $\small L+\epsilon :$ is the actual distance to the center of the pendulum

• Experiment if satisfactorily performed.