Measurements and Experimentation — Question 6

Given,

2 oscillations in 5 seconds, so

$\text {frequency per second} = \dfrac{2}{5} \\[0.5em] \Rightarrow \text {frequency per second} = \text {0.4 hertz} \\[0.5em]$

Hence, frequency of oscillation = 0.4 hertz.

As we know that,

$\text T = \dfrac{1}{\text f}$

So, substituting the value of f = 0.4 hertz, in equation above we get,

$\text T = \dfrac{1}{0.4} \\[0.5em] \Rightarrow \text T = 2.5\ \text s \\[0.5em]$

Hence, time period of pendulum is 2.5 s

(b) As we know,

$\text T = 2 \text π \sqrt{\dfrac{\text l}{\text g}} \\[0.5em]$

Given,

g = 9.8 ms^-2

and we know,

π = 3.14

T = 2.5 s

Substituting the values in the formula above we get,

$2.5 = 2 \times 3.14 \sqrt{\dfrac{\text l}{9.8}} \\[0.5em] \Rightarrow (\dfrac{2.5}{2 \times 3.14}) = \sqrt{\dfrac{\text l}{9.8}} \\[0.5em] \Rightarrow (\dfrac{2.5}{2 \times 3.14})^2 = \dfrac{\text l}{9.8} \\[0.5em] \Rightarrow (\dfrac{2.5}{6.28})^2 = \dfrac{\text l}{9.8} \\[0.5em] \Rightarrow (0.398)^2 = \dfrac{\text l}{9.8} \\[0.5em] \Rightarrow 0.158 = \dfrac{\text l}{9.8} \\[0.5em] \Rightarrow \text l = 9.8 \times 0.158 \\[0.5em] \Rightarrow \text l = 1.55\ \text m$

Hence, length of a seconds’ pendulum = 1.55 m