Question

A simple pendulum is swinging back and forth through a small angle, its motion repeating every 1.15 is. How much longer should the pendulum be made in order to increase its period by 0.22 s?

Answer 1

Answer:

ΔL=0.14m

Explanation:

The period of oscillations of simple pendulum is

$T=2\pi \sqrt{\frac{L}{g} }$

Here L is length of pendulum and g is acceleration due to gravity

So the length of pendulum is:

$L=\frac{T^{2}g }{4 \pi^{2} }$

So increase in length is:

ΔL=L₂-L₁

$=\frac{T_{2}^{2}g }{4 \pi^{2} }-\frac{T_{1}^{2}g }{4 \pi^{2} }\\=\frac{g}{4 \pi^{2} }[T_2^{2} -T_1^{2} ]\\ =\frac{9.8}{4 \pi^{2}}[(1.15+0.22)^{2}-(1.15)^{2} ] \\=0.14m$

ΔL=0.14m