Question

A simple pendulum of length \({}'L'\) and mass of the bob \({}'m'\) is oscillating with amplitude \({}'A'\). The maximum tension in the string is (\(g\) = acceleration due to gravity)

• A. \(\left[mg-\left(\frac{A}{L}\right)^{2}\right]\)
• B. \(\left[mg+\left(\frac{A}{L}\right)^{2}\right]\)
• C. \(mg\left[1+\left(\frac{A}{L}\right)^{2}\right]\)
• D. \(mg\left[1-\left(\frac{A}{L}\right)^{2}\right]\)

Hint:

Maximum tension occurs at the mean position, where T equals mg plus m v squared over L, and speed follows from energy conservation for small angles.

Answer 1

The Correct Option is C

Step 1: Use SHM speed
The maximum speed is $v_{\max}=\omega A=\sqrt{\frac{g}{L}}\,A$, so $v_{\max}^{2}=\frac{gA^{2}}{L}$.

Step 2: Forces at the bottom
$T_{\max}=mg+\frac{mv_{\max}^{2}}{L}$.

Step 3: Substitute
$T_{\max}=mg+\frac{mgA^{2}}{L^{2}}=mg\left[1+\left(\frac{A}{L}\right)^{2}\right]$, option (C).

Final Answer:
The maximum tension equals mg times one plus A over L squared. \[ \boxed{\text{(C) }mg\left[1+\left(\dfrac{A}{L}\right)^{2}\right]} \]