Question

A simple pendulum with a bob of mass m swings with an angular amplitude of 60 degree when its angular displacement is 30degree . What is the tension of the string

Answer 1

vo=√2g1/2[√3-1]

t-mgcos30deg=mv2/l

t-√3mg/2=mgl/l[√3-1]

t=mg[3√3-1/2

may b hlpfl

Answer 2

Answer:

The tension in the string will be $mg(\sqrt{3}-\frac{1}{2} )$

Explanation:

Let I be the length of the string.

At 60° height=lcos60=$\frac{I}{2}$

At 30° the height =lcos30=$\frac{\sqrt{3I}}{2} }$

​​Change in height=$\frac{\sqrt{3I}}{2} }-\frac{I}{2}$ =$\frac{1}{2}(\sqrt{3}-1 )$

Change in potential energy=$mg\frac{1}{2}(\sqrt{3}-1 )$

So change in potential energy will be equal to change in K.E since at the extreme position KI is zero

∴ The loss of KE will be due to the loss in PE

∴$\frac{1}{2}mv^{2}=mg\frac{1}{2}(\sqrt{3}-1 )$

∴$v^{2}=Ig\sqrt{3} -1$

The centrifugal force is given by

$mg\sqrt{3}-1$

Weight of mss=mgsin30°=$\frac{mg}{2}$

∴ Tension in the string will be equal to the sum of two forces

∴T=$mg(\sqrt{3}-\frac{1}{2} )$

Therefore, the tension in the string will be $mg(\sqrt{3}-\frac{1}{2} )$