11 questions · timed · auto-graded

Let equivalent spring constant be K4
$\therefore\frac{1}{\text{K}_4}=\frac{1}{\text{K}_2}+\frac{1}{\text{K}_3}=\frac{\text{K}_2+\text{K}_3}{\text{K}_2\text{K}_3}$
$\Rightarrow\text{K}_4=\frac{\text{K}_2\text{K}_3}{\text{K}_2+\text{K}_3}$
Now K4 and K1 are in parallel.
So equivalent spring constant $\text{K}=\text{K}_1+\text{K}_4=\frac{\text{K}_2\text{K}_3}{\text{K}_2+\text{K}_3}+\text{K}_1$
$=\frac{\text{K}_2\text{K}_3+\text{K}_1\text{K}_2+\text{K}_1\text{K}_3}{\text{K}_2+\text{K}_3}$
$\therefore\text{T}=2\pi\sqrt{\frac{\text{M}}{\text{k}}}=2\pi\sqrt{\frac{\text{M(k}_2+\text{k}_3)}{\text{k}_2\text{k}_3+\text{k}_1\text{k}_2+\text{k}_1\text{k}_3}}$

$\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$
The time period of the pendulum clock depends upon the acceleration due to gravity. As the earth-satellite is a free falling body and its g (effective acceleration due to gravity) is zero at the satellite, the time period of the clock is infinite.