Question

A simple pendulum with length 100 cm and bob of mass 250 g is executing S.H.M. of amplitude 10cm. The maximum tension in the string is found to be \(\frac{x}{40}\) N. The value of x is___.

Answer 1

Correct Answer: 99

For pendulum: \[ T_{\text{max}} = mg + \frac{mv^2}{L} \tag{1} \] Given \( m = \frac{1}{4} \, \text{kg}, L = 1 \, \text{m}, g = 9.8 \, \text{m/s}^2 \) and amplitude \( A = \frac{1}{10} \, \text{m} \) For SHM, \( K_{\text{max}} = \frac{1}{2} mv^2 = \frac{1}{2} m \omega^2 A^2 \) Using \( \omega = \sqrt{\frac{g}{L}} \), we get: \[ mv^2 = m \left( \frac{g}{L} \right) A^2 = m g \frac{A^2}{L} \] Substitute this into equation (1): \[ T_{\text{max}} = 2mg + \frac{mgA^2}{L^2} \] Substituting the given values: \[ T_{\text{max}} = 2 \times \frac{1}{4} \times 9.8 + \frac{1}{4} \times 9.8 \times \frac{101}{100} \] \[ T_{\text{max}} = \frac{98.98}{40} \] Therefore, \( x = 99 \)