Question

The length of the pendulum is 0.75 m. Time taken to complete 20 oscillations is 40s. Calculate its time period and (l/T^2) value​

Answer 1

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Answer 2

Given :

• Length of the pendulum (L) = 0.75m
• Time for 20 oscillations = 40s

To Find :

• Time period for one oscillation
• calculate $(l / \sf T^2)$

Basic Concepts :

Oscillation : A complete to and fro motion of the Bob of the pendulum is an oscillation.

Period : The time taken by the pendulum to complete one oscsilation is called Time period (T)

Solution :

Time for one oscillation = 40s

Time for one oscillation (T) =$\sf \dfrac{40}{2}\Rightarrow 2s$

Directly substitute the values to find $(l / \sf T^2)$

$\longrightarrow{\dfrac{l}{\sf T^2}}\\\\\longrightarrow {\sf \dfrac{0.75}{(2)^2}}\\\\\longrightarrow {\sf \dfrac{0.75}{4}}\\\\\longrightarrow {\sf 0.1875m^{-2}}$

Required Answer :

Time period for one oscillation is $\underline{\sf 2s}$

The value of $(l / \sf T^2)$ is $\underline{\sf 0.1875m^{-2}}$