Question

A simple pendulum has a length L and a bob of mass m. The bob is oscillating with amplitude A. Show that the maximum tension (T) in the string is (from small angular displacement) Tmax = mg [1 + (A/L)²]

Answer 1

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)

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Answer 2

Explanation:

We have to prove :

$T_{max} = mg [1 + (\dfrac aL)^2] ...(1)$

at mean position of pendulum

we know that

v = a × ω ... (2)

where a = length of the pendulum

$w = \sqrt {\dfrac gL}$

Now put the value of w in equation (2)

$v = a \times \sqrt\dfrac{g}{L}$

Now, put the value of v in equation (1)

$T_{max} = mg + \dfrac{m \left(a \sqrt {\dfrac gL}\right)^2} {L}\\\\T_{max} = mg + \dfrac {mga^2}{L^2}\\\\T_{max} = mg(1 + \dfrac {a^2}{L^2})$

Hence proved.