A simple pendulum with length L and mass M of the bob is vibrating with amplitude a. Then the maximum tension in the string is

Q. A simple pendulum with length L and mass M of the bob is vibrating with amplitude a. Then the maximum tension in the string is

  1. Mg
  2. Mg \left[ 1 + \left( \frac{a}{L}  \right)^2 \right]
  3. Mg \left[ 1 + \frac{a}{L} \right]^2
  4. Mg \left[ 1 + \frac{a}{2L} \right]^2

Solution

For bob to move in a circular path, the tension is maximum at the lower point. (i.e. at mean position). The maximum tension on the string can be calculated as follow:

    \[T_{max} - mg = \frac{mv^2}{r}\]

But r = L (the length of the pendulum)

(1)   \begin{equation*} T_{max} = mg + \frac{mv^2}{L}\end{equation*}

But v=\omega r which means v = \omega a since, amplitude is ‘a’ given by a question.

And in SHM, \omega = \sqrt{\frac{g}{L}}

So,

(2)   \begin{equation*}  v = \sqrt{\frac{g}{L}}. a\end{equation*}

From equation 1 and 2, we can write:

    \[ T_{max}=mg + \frac{m\frac{g}{L}.a^2}{L} \]

    \[T_{max} = mg + \frac{mga^2}{L^2}\]

    \[T_{max} = mg \left[ 1+ \frac{a^2}{L^2}\right]\]

    \[ \boxed{T_{max} = Mg \left[ 1 + \left( \frac{a}{L}  \right)^2 \right]} \]

Ans: (b) Mg \left[ 1 + \left( \frac{a}{L}  \right)^2 \right]

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