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

Answer:

The new period will be reduced by 50%

Explanation:

The period of pendulum is given by;

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

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

L₂ = 25/100(L₁)

L₂ = 0.25L₁

$\frac{L_1}{T_1^2} = \frac{L_2}{T_2^2}\\\\T_2^2 = \frac{T_1^2L_2}{L_1} \\\\T_N^2 = \frac{T^2(0.25L_1)}{L_1}\\\\ T_N^2 =0.25T^2\\\\T_N = \sqrt{0.25T^2}}\\\\T_N = 0.5 T$

Thus, the new period will be reduced by 50%