Question

(b) If you decrease the length of the pendulum by 25%, how does the new period TN compare to the old period T?

Answer 1

The new period will be reduced by 50%

Step-by-step explanation:

The period of pendulum is given by;

`T= 2\pi\sqrt{(L)/(g) }\\\\(T)/(2\pi) = \sqrt{(L)/(g) }\\\\((T)/(2\pi) )^2 = {(L)/(g)}\\\\(T^2)/(4\pi ^2) = {(L)/(g)}\\\\T^2((g)/(4\pi ^2)) = L\\\\ (g)/(4\pi ^2)= (L)/(T^2)\\\\(L_1)/(T_1^2) = (L_2)/(T_2^2)`

When the length is decreased by 25%, the new length L₂ is given by;

L₂ = 25/100(L₁)

L₂ = 0.25L₁

`(L_1)/(T_1^2) = (L_2)/(T_2^2)\\\\T_2^2 = (T_1^2L_2)/(L_1) \\\\T_N^2 = (T^2(0.25L_1))/(L_1)\\\\ T_N^2 =0.25T^2\\\\T_N = √(0.25T^2)}\\\\T_N = 0.5 T`

Thus, the new period will be reduced by 50%