Question

A simple pendulum, consisting of a small ball of mass m attached to a massless string hanging vertically from ceiling is oscillating with an amplitude such that the maximum tension in the string is related to the minimum tension by tmax = 2 tmin. What is the value of the maximum tension in the string?

Answer 1

Answer:

Tmax = mg + mv²/l

At mean position

V = a * w and w =$\sqrt(g/l)$

v = a $\sqrt(g/l)$

Hence

Tmax = mg ( 1 + a²/l²)

by this posses you get it answer

Answer 2

Answer:

The value of the maximum tension in the string will be $\frac{3mg}{2}$.

Explanation:

By applying the law of conservation of energy at extreme and mean positions we have,

$\frac{1}{2}mv^2=mgl(1-cos\theta)$

$v^2=2gl(1-cos\theta)$          (1)

Where,

m=mass of the pendulum

v=velocity at the mean position

l=length of the string

The maximum tension at the mean position is given as,

$T_m_a_x=mg+\frac{mv^2}{l}$              (2)

By substituting equation (1) in equation (2) we get;

$T_m_a_x=mg+\frac{m(2gl(1-cos\theta))}{l}$

$T_m_a_x=mg+2mg(1-cos\theta)$             (3)

The minimum tension at the extreme position is given as,

$T_m_i_n=mgcos\theta$              (4)

From the question we have;

$T_m_a_x=2T_m_i_m$            (5)

By substituting equations (3) and (4) in equation (5) we get;

$mg+2mg(1-cos\theta)=2mgcos\theta$

$3mg =4mgcos\theta$

$cos\theta=\frac{3}{4}$             (6)

By substituting the value of cosθ in the equation (1) we get;

$v^2=2gl(1-\frac{3}{4})=\frac{gl}{2}$          (7)

By putting the value of v² in equation (1) we get;

$T_m_a_x=mg+\frac{m}{l}\times \frac{gl}{2}$

$T_m_a_x=\frac{3mg}{2}$

Hence, the value of the maximum tension in the string will be $\frac{3mg}{2}$.